Handbook
- A Comprehensive Guide to Placements
Preface
Campus placements play an important role in a students’ life. It sometimes becomes a turning point in their lives too. As in the case with every worthy endeavor, preparation makes a lot of difference towards achieving it. This book is primarily targeted at students looking for a guide to help them in preparation for placements.Target Audience
This book is primarily targeted at graduates and post-graduates of computer science (and related disciplines), who want to refresh and improve their knowledge in core computer science areas focusing towards placements.
This book also serves as a question bank for Computer Science. However this book has a three main differences from other question banks available in the market. This book covers only the most important of the subjects; within the concerned subject it covers only essential questions and most in cases, it gives detailed answers. So, this book stands with an advantage over the question banks.
This book can also serve for students appearing for various competitive exams related to computer science. For instance the first part of book is suitable for the entrance exam preparation for MCA for various universities and colleges throughout the country. Similarly this book can be handy for the students appearing for NCST’s CST (Competency in Software Technology) Exams of levels E, I, D, G and DOEACC’s exams and other similar competitive exams.
This book is very handy for students and teachers looking for a book that cover the essence of important subjects in form of question and answers.
Part II – Concepts of this book can serve also as a FAQ on important topics in Computer Science.
Anyone interested in increasing the aptitude in computer related areas, can use this book. Even though this book is not written having beginners in mind, with little struggle, they can make-up to read the book, because this book addresses main concepts from the basics moving towards more tough areas.
Sometimes it can also serve as a general reference; when in doubt about important topic, they can always look for information about that topic.
This book neither has full-fledged/in-depth coverage of subjects, nor can be used as teaching material/textbook. If that is your intention, there are other books available; few of them listed under ‘suggested reading’. This book is meant to be far from them. It is rather a refresher of core concepts in important subjects.
Each chapter is organized such that the reader can start from any chapter and know more about that topic and is mostly independent of other chapters. General readers may be selective in reading the chapters according to their need.
The book starts with a short introduction to placements. The whole book is divided into two parts:
Part I - Aptitude and
Part II - Concepts.
‘Part I - Aptitude’ covers the aptitude part essential for appearing for the written tests for the various companies and exams. This part mostly follows a question-answer-explanation format, because for aptitude preparation, the objective type is not sufficient. If the reader is able to prepare and answers in this manner, answering objective questions is direct and easy.
‘Part II – Concepts’ is for information on areas of specialization that will be useful for attending technical interviews (and sometimes asked in the written tests too). It is more likely that the interviewers ask questions on most important of the topics in Computer Science to judge the technical capability of the candidate. This part serves to be a refresher of concepts on important topics and covers most of the important underlying facts/concepts that a computer science student should be familiar with.
The ‘Part I – Aptitude’ has a problem solving approach and consists of four chapters. We encourage the readers to try to solve the problems provided than to just read them. Problem solving is an area where there are more possibilities for making silly errors; only practice can help in avoiding them. This will also enable them to access their strong-points and weaknesses and help improving them. Since explanations are provided, it will help understanding the concept behind the question asked and reason it out why the answer is.
Chapter 1 is on C Aptitude. This is the most important of the aptitude to be developed. A careful selection is made in picking and framing the questions, to enable coverage of all topics of importance in C, and their complexity vary from very easy to tough.
The next chapter is on C++ and Java aptitude. C++ is such a complex language that it is almost impossible to cover all the areas, so we have given our best to address most important of the concepts. Java is a relatively new language and it will take some more time for the companies to concentrate more on Java. So there are comparatively few questions on Java and that too are intended to be sampling questions on most of the must-know topics in Java.
The third chapter is on UNIX and SQL. This part is useful both for written tests and interviews; so it doesn't follow the question-answer-explanation format. UNIX covers the most important of the usage of various commands in the common subset of UNIX systems. The shell programming is not covered because there exits many popular UNIX shells and providing a common subset of questions for all shells is not possible. SQL questions are targeted at improving the ability to solve commonly asked SQL queries given a schema. PL/SQL is not covered for the same reason as with shell programming.
The final chapter of the first part of this book is on general aptitude. Quantitative aptitude takes major part in placement selection papers. It is almost impossible to cover the general aptitude section in a full-fledged manner. So important formulae are listed in the beginning of the chapter. This chapter is just gives a taste of how the general aptitude questions will look like in actual written tests.
The ‘Part – II – Concepts’ is made-up of 7 chapters.
The chapter on Data Structures and Algorithms is of primary importance of all the chapters in this part. Since this subject is a base subject for computer science, a clear understanding of the concepts is necessary and this chapter serves the purpose of covering the important of the concepts to be understood.
The next chapter is on Database Management Systems. This chapter covers most of the important rules and concepts that an interviewer may expect that the candidate be aware of.
The next chapter is on Object Oriented Programming and Object Oriented Analysis and Design. OOPs concepts are covered based on C++ since it is the widely used language. OOAD addresses the analysis and design concepts in general and UML in particular.
The next two chapters, Operating Systems and UNIX are interrelated. The general concepts on OS are covered in the chapter on OS and the application of those concepts of how they take shape in the UNIX operating system is discussed in the chapter on UNIX. The chapter on UNIX has three sub-sections: process management, memory management and file management. Some questions that are specific to UNIX System V are there in section of memory management and such contexts are specified explicitly.
Computer Networks is a hot topic in today’s context and many of the definitions are covered in this chapter. The final chapter is on client/server computing and there is more possibility that the interviewers will ask questions on this since this is the age of Internet
As a whole the chapters are organized such that they provide an overall coverage of most of the concepts required to be known by a student attending for placements.
.
Appendix-I of the book lists the subjects covered for the written tests conduced by important IT companies. Appendix-II lists the books suggested for further reading on selected topics.
Notes on Formatting
The questions are italicized to distinguish them from the answers. The words that are italicized inside the normal text indicate special meaning or their importance.
TABLE OF CONTENTS
Preface
An Introduction to Placements
PART – I
APTITUDE
C Aptitude
C++ and Java Aptitude
UNIX Commands and SQL Statements
General Aptitude
PART – II
CONCEPTS
Data Structures and Algorithms
RDBMS concepts
OOPs and OOAD Concepts
Operating System Concepts
UNIX Concepts
Computer Networks
Client/Server Computing
APPENDIX – I
List of Subjects Covered in Written Tests by Various Companies
APPENDIX – II
Suggested Reading
Introduction
This chapter gives an overview on campus placements (particularly on-campus, still the most of the ideas will apply for off-campus selections too) and preparations for the placements. You might already know most of the information given here, but we think there is no harm in giving it.
What Companies Expect
The IT field is a constantly growing one, where the technology of yesterday is a relic today. There is a wide gap between the requirement and supply of the skilled manpower. So there is a very high demand for such capable people. They are looking forward for such people and expect their employees to be dynamic enough to have the aptitude for continuous learning and a willingness to work more. Software development is teamwork, so the companies expect reasonably good communication skills. It’s enough that you have the capability to express your thoughts freely.
The Selection Procedure
Most of the companies follow the similar methodology for campus selections. We feel that the companies adopt more a filtering approach to selection strategy in initial stages.
Many companies mandate a 60%-75% average in the qualifying degree. Few companies look for good track record from school days itself. Consistent performance is also sometimes taken into account: some companies require that the candidates should have no arrears.
Next comes the written test. The test topics and weight age given for each of the subject tested varies between the companies. The aim is to select the candidates with good aptitude that includes C aptitude and general aptitude (analytical aptitude and reasoning). Depending on the companies the other subjects tested varies. The list of companies and the subjects they cover for written test in given in Appendix-I of this book.
When you are selected in the written test, some companies have Group Discussion for further filtering of the candidates.
Finally you have personal interview. Depending on the company there may be one or two interviews; in case of technical session and HR/Stress interviews. Performance in interviews is crucial in the selection process because it is where you are evaluated fully and final decision is made on your selection.
What to prepare and How to prepare
Proper planning for placements will help a lot in improving your chances for getting placed. Identify your strong points and weaknesses and plan accordingly. Preparation should start well ahead of the placement dates, and 6 months time will be more than enough.
There are many facets that are to be considered for preparing for placements. The preparation should constitute: improving technical capabilities, increasing the general aptitude, developing good communication skills and mental preparation.
Technical Capabilities:
Long before placements start, you should have clearly identified your areas of interest and specialization. There is more possibility that the interviewers will ask questions in those areas.
As far as technical capabilities are concerned, it’s more a long-term preparation. You can go through the textbooks of the concerned subjects once again and know the subject better. Don't mug-up; this is not for your semester exams. Concentrate more on concepts than data; employees prefer good thinkers to RAMs.
A wide knowledge in all the fields concerned with computer science is a very big advantage. This book can help very much as a refresher of important concepts and having an overall view of almost all the important subjects concerned with computer science in placements point of view. For full coverage of information you should read the text-books and few of the books on selected topics are listed in Appendix-II “suggested reading”.
Data Structures and Algorithms, Computer Networks, Operating Systems, DBMS are very important areas in Computer Science. Knowledge in these areas will be a big advantage, even if your area of interest/specialization doesn't happen to be one of these subjects.
Non-computer science students need to be strong in their areas of graduation. Still knowledge of computers and C language is very much desirable.
If you don't know C language basics, start learning C immediately. Knowledge of C language is a must and almost all the companies test C aptitude. C++, Java and UNIX aptitude will be an added advantage. It is worthwhile investing lot of time improving your aptitude in these areas; they are here to stay in IT field for a long time.
General Aptitude:
Good skills in general aptitude is a must to pass the written test, and sometimes in interview also general aptitude skills are tested. You can perform better if you prepare well for improving this aptitude. Both short and long term preparation shall be useful. Know the shortcuts for answering these questions since time is an important factor in answering these questions. Memorize important formulae and many of them are listed in the chapter dedicated for general aptitude in this book.
Now lets look at few tips on solving the general aptitude questions.
- Each one of you will have your own positive and negative aspects in problem solving. First identify your gray areas first and work on them assiduously.
- The two most important attributes when we go for any aptitude test is speed and accuracy. One without the other is meaningless.
- Remember that it is almost humanly impossible to solve all the problems in the stipulated time because it will then reflect poorly on the standard of the question paper being set and no company will ever allow that. So don't be in too much of a hurry to answer all the questions and in the process make some stupid errors.
- Reasoning ability is more important and not our mathematical skills. Almost 95% of our problems do not require too much of a mathematical intellect on our part.
- When you sit in for a test try to identify those problems that take more time to solve. Remember that all the questions carry equal marks and it doesn't make sense for us to solve one problem in say 30 seconds and another in, say 3 minutes. Skip such questions in the beginning and come back to them later if and when you have time.
Communication Skills:
Having good oral communication skills is a must for success in the personal interview. It’s enough that you are able to communicate your ideas effectively without any inhibitions in English.
English is a language in which you can achieve a considerable fluency in a short time span. This requires a conscious effort on your part, and thus, a good command over the English language can be achieved. Make it a point that you speak in English with your friends and ask them to point out the mistakes you commit while you speak.
Mental Preparation
Mental preparation plays a vital role in the placements. We should have the confidence that we can get placed. This is the basic and most important point to be noted. Build a positive self-image and project the best in you.
Conclusion
The problem India faces is not the problem of unemployment; it is rather the problem of 'unemployability'. We should make ourselves capable and improve ourselves to make suitable for being employed. For that preparations will help a lot.
 |
Note: All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Ø Programs run under DOS environment,
Ø The underlying machine is an x86 system,
Ø Program is compiled using Turbo C/C++ compiler.
Ø Proper and required header files are included,
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).
Predict the output or error(s) for the following:
1. void main(){
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
2. main(){
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
3. main(){
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precision with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .
4. main() {
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.
1. main(){
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.
2. main(){
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .
3. main(){
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.
4. main(){
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.
5. main(){
int i=3;
switch(i) {
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.
6. main(){
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
7. main(){
char string[]="Hello World";
display(string);
}
void display(char *string){
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
8. main(){
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
9. #define int char
main(){
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
10. main(){
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14, NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
11. main(){
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77 (77 is the ASCII value for "M");
So we get the output 77.
12. main(){
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
13. main(){
struct xx{
char name[]="hello";
};
struct xx *s;
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in structure declaration.
14. main(){
struct xx{
int x;
struct yy{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
No output.
Explanation:
Pointer to the same type of structures are known as self referential structures. They are particularly used in implementing datastructures like trees. Structures within structures are known as nested structures.
15. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
16. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.
17. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
The macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64
18. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
Ø *p that is value at the location currently pointed by p will be taken
Ø ++*p the retrieved value will be incremented
Ø when ; is encountered the location will be incremented, that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.
19. #define a 10
main()
{
#define a 50
printf("%d",;
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.
20. #define clrscr() 100
main(){
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem.
21. main(){
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.
22. main(){
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).
23. enum colors {BLACK,BLUE,GREEN}
main(){
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
24. void main(){
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
The second pointer is of char type and not a far pointer
25. main(){
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.
26. main(){
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.
27. main(){
int i=1;
while (i<=5){
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun(){
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.
28. main(){
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
29. void main(){
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i.
30. void main(){
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
31. main(){
int i=1,j=2;
switch(i){
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
32. main(){
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read.Here 10 is given as input which should have been scanned successfully. So number of items read is 1.
33. #define f(g,g2) g##g2
main(){
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
34. main(){
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
Before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).
35. main(){
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.
36. main(){
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.
37. main(){
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
38. main(){
show();
}
void show(){
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
1. main( ){
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***;
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
| 2 | 4 | 7 | 8 | 3 | 4 | 2 | 2 | 2 | 3 | 3 | 4 |
100 102 104 106 108 110 112 114 116 118 120 122
Thus, for the first printf statement a, *a, **a give address of first element. Since the indirection ***a gives the value. Hence, the first line of the output. For the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.
2. main( ){
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++){
printf(“%d” ,*;
a++;
}
p = a;
for(j=0; j<5; j++){
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
3. main( ){
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
| 0 | 1 | 2 | 3 | 4 |
100 102 104 106 108
p
| 100 | 102 | 104 | 106 | 108 |
1000 1002 1004 1006 1008
ptr
| 1000 |
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.
4. main( ){
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as
M | O | U | S | E | \0 |
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.
M | T | R | A | C | K | \0 |
The third input starts filling from the location 102
M | T | V | I | R | T | U | A | L | \0 |
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
5. main( ){
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.
6. main ( ){
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.
7. main(){
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i){
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.
8. int i,j;
for(i=0;i<=10;i++){
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion is a good debugging tool to make use of.
9. main(){
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).
10. What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
11. What will be the position of the file marker?
a) fseek(ptr,0,SEEK_SET);
b) fseek(ptr,0,SEEK_CUR);
Answer :
a) The SEEK_SET sets the file position marker to the starting of the file.
b) The SEEK_CUR sets the file position marker to the current position of the file.
12. main(){
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.
13. What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF) ;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.
14. main(){
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.
15. main(){
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.
16. main() {
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.
17. main(){
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.
18. #define FALSE -1
#define TRUE 1
#define NULL 0
main(){
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.
19. main(){
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".
20. main(){
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
21. #define max 5
#define int arr1[max]
main(){
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.
22. int i=10;
main() {
extern int i; {
int i=20;{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as, const volatile unsigned which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.
23. main(){
int *j;{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.
24. main(){
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.
25. main() {
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
1. main(){
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.
2. main() {
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.
3. main(){
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
The expression i+++j is treated as (i++ + j).
4. struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main(){
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
5. struct point{
int x;
int y;
};
struct point origin,*pp;
main(){
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
6. main(){
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i){
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.
7. main(){
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.
8. main(){
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z){
return z-32;
}
Answer:
Compiler error
Explanation:
Declaration of convert and format of getc() are wrong.
9. main(int argc, char **argv){
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)int num1,num2;{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.
10. int one_d[]={1,2,3};
main(){
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
11. aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main(){
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.
12. main(){
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.
13. main(){
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.
14. main(){
int i;
i = abc();
printf("%d",i);
}
abc(){
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.
15. int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution will be,
t i x
4 0 -4
3 1 -2
2 2 0
16. main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.
17. main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.
85. In the following pgm add a stmt in the function fun such that the address of 'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
86. What are the following notations of defining functions known as?
i. int abc(int a,float b) {
/* some code */
}
ii. int abc(a,b)
int a; float b; {
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation
87. main(){
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.
88. main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(;
}
abc(char a[]){
a++;
printf("%c",*;
a++;
printf("%c",*;
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.
89. func(a,b)
int a,b;{
return( a= (a==b) );
}
main(){
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.
90.void main(){
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.
91.void main(){
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret){
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.
92.void main(){
char a[]="12345\0";
int i=strlen(;
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.
93.void main(){
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
94.void main(){
int i;
char a[]="\0";
if(printf("%s\n",)
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.
95.void main(){
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of time.
96.void main(){
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its declaration.
So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).
97.void main(){
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
85.void main(){
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.
86.main(){
unsigned int i=10;
while(i-->=0)
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will always be true, leading to an infinite loop.
87.main(){
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
88.main(){
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
89.main(){
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.
90.main(){
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.
91.main(){
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.Bit-wise operators and % operators cannot be applied on float values.fmod() is to find the modulus values for floats as % operator is for ints.
92.main(){
int i=10;
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k){
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.
93.void pascal f(int i,int j,int k){
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k){
printf(“%d %d %d”,i, j, k);
}
main(){
int i=10;
printf(" %d\n",i);
i=10;
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.
94.What is the output of the program given below
main(){
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
-128
Explanation:
Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.
95.main(){
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
infinite loop
Explanation:
The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.
96.main(){
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.
97.Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer:
Definition. x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.
98.What is the output for the program given below?
typedef enum errorType{warning, error, exception,}error;
main() {
error g1;
g1=1;
printf("%d",g1);
}
Answer:
Compiler error: Multiple declaration for error
Explanation:
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note:
The extra comma in the declaration,enum errorType{warning, error, exception,} is not an error. An extra comma is valid and is provided just for programmer’s convenience.
99.typedef struct error{int warning, error, exception;}error;
main(){
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer:
1
Explanation:
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.
Note:
This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!
100. #ifdef something
int some=0;
#endif
main(){
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
101. #if something == 0
int some=0;
#endif
main(){
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
0 0
Explanation:
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.
102. What is the output for the following program?
main(){
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer:
1
Explanation:
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays. arr2D is made up of a 3 single arrays that contains 3 integers each .
|
|
| ||
|
The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is printed.
103. void main(){
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer:
You can answer this if you know how values are represented in memory
Explanation:
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.
104. int swap(int *a,int *b){
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main() {
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer:
x = 20 y = 10
Explanation:
This is one way of swapping two values. Simple checking will help understand this.
105. main(){
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
106. main() {
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.
107. main(){
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.
108.
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main(){
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing "bye".
109. main(){
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.
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