C Interview Questions and Answers _ TechInterviews

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Answer: 10, 5

10, 5

What will be printed as the result of the operation below:

main()

{

char *ptr = Cisco Systems;

*ptr++; printf(%sn,ptr);

ptr++;

printf(%sn,ptr);

}

Answer:Cisco Systems

isco systems

5.


main()

{

char s1[]=Cisco;

char s2[]= systems;

printf(%s,s1);

}

Answer: Cisco

6.


main(){

char *p1;

char *p2;

p1=(char *)malloc(25);

p2=(char *)malloc(25);

strcpy(p1,Cisco);

strcpy(p2,systems);

strcat(p1,p2);

printf(%s,p1);

}

Answer: Ciscosystems

7.

The following variable is available in file1.c, who can access it?:

static int average;

Answer: all the functions in the file1.c can access the variable.

8.

WHat will be the result of the following code?

#define TRUE 0 // some code

while(TRUE)

{

// some code

}

Answer: This will not go into the loop as TRUE is defined as 0.

9.


int x;

int modifyvalue()

{

return(x+=10);

}

int changevalue(int x)

{

return(x+=1);

}

voidmain()

{

int x=10;x++;

changevalue(x);

x++;

modifyvalue();

printf("First output:%dn",x);

10.

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13 30-09-2012 17:19


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x++;

changevalue(x);

printf("Second output:%dn",x);

modifyvalue();

printf("Third output:%dn",x);

}

Answer: 12 , 13 , 13


main(){

int x=10, y=15;

x = x++;

y = ++y;

printf(%d %dn,x,y);

}

Answer: 11, 16

11.


main()

{

int a=0;

if(a==0)

printf(Cisco Systemsn);printf(Cisco Systemsn);

}

Answer: Two lines with Cisco Systems will be printed.

12.

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98 COMMENTS ON C INTERVIEW QUESTIONS AND ANSWERS

moimart

Posted 4/7/2006 at 4:16 am |Permalink

#1 Its very very very easy. When printf, p2 points to the 4th position not the

beginning. (from 4th position to 20th position is 0)

Joe Yabyam

Posted 6/14/2006 at 4:46 pm | Permalink

#5 is correct. Both GCC/linux as well as Windows (both VC6 and Cygwin) print as

answered - (Answer:Cisco Systems

isco systems)

This is assuming you #include prior to main().

Joe Yabyam



13 30-09-2012 17:19


4/13

re: #5 is correct That is also if you fix the quotes problem in the strings.

Ankur

Posted 7/4/2006 at 9:12 a m | Permalink

there is some prob with q 5 and q 10

i use turbo c compiler

*ptr++ or *(ptr++)

if i nw print ptr as a charit should be printed as C..

but tc displays error code has no effect..

even with %s it says code has no effect.

with (*ptr)++ it is correct..

! is printed

please ans this query..

also in q no. 10 if i mention x as static

it should persist bw functn calls

but still 12,13,13 is output

tel me why?????????

srikPosted 7/8/2006 at 2:52 pm | Permalink

for #11 it the answer is 10 16.

I got scared for a while there.

sreejesh

Posted 7/21/2006 at 5:28 am | Permalink

can we multiply two long integers by using only integer and string

Scott


Question 8This fails to compile in gcc:

void test()

{

stvar = 3;

}

static int stvar = 3;

Its a small point, but that seems to be what this test is about.

navya


#10 explanation

1)variable X is intialized to 10

2)x++ will have the value 11

eventhough the function changevalue() have a return stmt there is no variable to store

the value

3) after excecution of the function changevalue() the control comes back to the next

stmt x++ where the value of x becomes 12

4)the modifyvalue() function also dont store the value(similar to the above said

changevalue())

5)so the value of x ie 12 is printed

rest of the excecution is similar please try it out

Divya

Posted 9/13/2006 at 2:53 pm |Permalink


13 30-09-2012 17:19


5/13

hi,

what wil be the output of below function return and whats the importance of volatile

here

int sqrt(volatile int *ptr)

{

return (*ptr**ptr);

}

ben lev


Pretty good questions.

Only one error in your output in question #4

Your output

10, 5

10, 5

should be

10 5

10 5

No comma between the numbers only space.

Thanks

Wei


Explanation to #10:

First line x is global variable, the x in main() is local variable. The fu nction

modifyvalue() only effects global x, but not the local x in main().

As to changevalue(x), its called by value, the x in function changevalue(x) wont effect

the x in main().

bharat biswal


How can you find the number of elements stored in a static array at a time ?

mannu


hi everybody

main()

{

int x=10, y=15;x = x++;

y = ++y;

printf(%d %d\n,x,y);

}

ans should be

11,15

for x++ x=x+1 i.e 11

y;but due to preincrement value of y after print is 15

can u clarify the ans

Surya


helloo manu ,

as u said ys value shud be 15 is correct but u should note the line as it is y=++y which


13 30-09-2012 17:19


6/13

means the value will be incremented and stored into y itself.

So when it come to next line it will show as 16 only.

mrinu


hi

++y

andy=++y;

both r one and the same.using the second one that is y=++y is meaningless becau se

++y itself means increment y and store it back in y itself so using y=++y is worth

nothing.

and because there r no expresions used ++y and y++ both give the same answer,

ajay kant singh


In Question 2, at step 1: x=y++ + x++

it 20+35=55and at second statement , ++y + ++x

and 22+56=78

the answrer is :5578

can you explaini it to me..

Anders


Question #3:

This is a peculiarity with printf if I am not mistakensince a separate shift with print

produces the sequence 5 20 5. The printf version is correctly 5 20 1, so if anyone can

explain the printf mangling of the 3rd bit, thank you !

malaram


Hi Maxmelbin

The answer of Q#3 is 5,20,1

because

a>b=a/(square of b)

That means

for a=5a>2=5/(2*2)=5/4=1(take only integer value)

Jessie


Hi!

I dont understand the 10th question.Can any one of u explain me?


int x;

int modifyvalue(){

return(x+=10);

}

int changevalue(int x)

{


13 30-09-2012 17:19


7/13

return(x+=1);

}

void main()

{

int x=10;

x++;

changevalue(x);

x++;

modifyvalue();

printf(First output:%d\n,x);

x++;

changevalue(x);

printf(Second output:%d\n,x);

modifyvalue();

printf(Third output:%d\n,x);

}

Answer: 12 , 13 , 13

sangeeta


the variable x been defined in main (which is local to main) can be accessed in main,

where as the variable x accessed in the subroutines is the global variable

sangeeta


in other words x in modifyvalue() is the global variable, in changevalue(x) is the x local

to changevalue, in main() again as it is defined here its local to main.

hope this clears Jessies doubt

Ratheesh


hi ajay kant singh ,

depends on compiler

Ashish


Hello All,

Can ne one help me in getting the right answer of the question?

q-> Which one of the following is NOT a default promotion?

Choice 1 If either operand of an arithmetic operator is unsigned long, the other

operand is promoted to unsigned long.

Choice 2 If either operand of an arithmetic operator is unsigned int, the other operand

is promoted to unsigned int.

Choice 3 If either operand of an arithmetic operator is int, the other operand is

promoted to int.

Choice 4 If either operand of an arithmetic operator is double, the other operand is

promoted to double.

Choice 5 If either operand of an arithmetic operator is short, the other operand is

promoted to

sirisha

Posted 3/9/2007 at 11: 12 pm | Permalink


13 30-09-2012 17:19


8/13

Reffered to Poly

The answer to the Q # 4 is correct.

The first function is declared as global and hence the x and y values are swapped after

calling the swap(x,y) function. Now the current values of x and y are 10 and 5.

When the second function swap1(x,y) is called the values x=10 and y=5 are passed to

the function. But as this is not call by reference the swapped values are lost once it

comes out of the man function. Hence the original values x=10 and y=5 are printed.

satya


priyanka said,

hi ppl ,

#2. x=20, y=35

now x= y++ + x++;

i.e. 35 + 20 ; // after this x= 21 ,y=36

now y= ++x + ++y ;

i.e. 22 + 37 ;

So the answer sd : 5559(i.e 55 and 59 ,,94 is out of question..Plz explain) .

This answer was wrong..

correct answer is..

x= y++ + x++;

i.e. 35 + 20 ; // after this x= 76 ,y=36

now y= ++x + ++y ;

i.e. 77 + 37 ; // after this x=77 , y=114;

is it clear

Sathish Kumar


main()

{

int x=20,y=35;

x=y++ + x++;

y= ++y + ++x;

printf(%d%d\n,x,y);

}

Answer : 5794

Can anybody explain me this increments in a order..

Bibin Thomas


Q:2

x=y++ + x++;

y= ++y + ++x;

Q:11

x = x++;

Lots of comments on this!!!

i = i++ is UB(Undefined Behaviour) beacuse youre modifying the same variable more

than once without an intervening sequence point.

For more info read: .

Bibin Thomas


(UB)


13 30-09-2012 17:19


9/13

for more info:

http://en.wikipedia.org/wiki/Sequence_point

Me


16 is incorrect (answer is not 11 16, but 10 16):

$ g++ test2.cpp ;./a10 16

$ g++ version

g++ (GCC) 3.4.4 (cygming special) (gdc 0.12, using dmd 0.125)

Copyright (C) 2004 Free Software Foundation, Inc.

James


Answer to muralipattas Qn,

==============================================================

1.if i hav a declatration like this int a[5]={1,3); why the elements a[2],a[3],a[4] areequal to zero.

2.int a[5];

a[0]=4;

a[1]=5;

then why the values a[2],a[3],a[4] are garbage.

==============================================================

In the first case, as explicit array size is specified, and a shorter initiliazation list is

specified, the unspecified elements are set to zero by the compiler. So you will get

zeroes for the 2,3,4 elements.

In the second case the array has no initialization, so the values are undefined during

compile time. So the array will have garbage values for all locations. And you are

changing the values of indexes 0 and 1 only at runtime.

RameshReddy


#

satya said,

priyanka said,

hi ppl ,

#2. x=20, y=35

now x= y++ + x++;

i.e. 35 + 20 ; // after this x= 21 ,y=36

now y= ++x + ++y ;

i.e. 22 + 37 ;

So the answer sd : 5559(i.e 55 and 59 ,,94 is out of question..Plz explain) .

This answer was wrong..

correct answer is..

x= y++ + x++;

i.e. 35 + 20 ; // after this x= 76 ,y=36

now y= ++x + ++y ;

i.e. 77 + 37 ; // after this x=77 , y=114;

is it clear

My opinion is like, Answer is compiler dependent.

It behaves different for different compilers.

In GCC its ans is 5693

Its Compiler dependent.


13 30-09-2012 17:19


10/13

sapkota


#

kauai said,

On #1,

while(*p2++ = *p1++); When you do an assignment like this isnt always

true? Hence an infinite loop (or at least until there is a memory access violation) ??

THIS IS TOTALL WRONG ,

since value which is pointed by p2 (ie name) is a string and string always terminat

with 0 so there is no chance at all that causes Access Violation.. Hope it clears

SHYJU KV


clarification for Question#2

step1:

x=y++ + x++; (note operator precedence ++ have high then +)

so x=21 + 36=57

step2:

y=++y + ++x; (note operator precedence all have same so precedence is right to left)

so y=36 + 58 =94 (++y value assigned only after addition )

Karthik ...


ya ,ans 4 2nd 1 is,

5794

x=20; y=35;

I step: x=y++ + x++;

x=36 + 21 = 57;

II step:y=++y + ++x;

y=36 + 58 = 94; /*here value of x assigns 57 then incremented to 58*/

Nw got it

So ans s 5794

Karthik ...


Can any1 say wat s d o/p of printf(%d);

Hi 2 all , i think d ans 4 d abve qn s

o/p shows some Garbage values ,its my guess,so any other answers / comments

priya


= y++ + x++;

i.e. 35 + 20 ; // after this x= 76 ,y=36

now y= ++x + ++y ;

i.e. 77 + 37 ; // after this x=77 , y=114;is it clear

explain again


f 13 30-09-2012 17:19


11/13

Biswajit Dash

Posted 9/6/2007 at 1 1:34 am |Permalink

int main()

{

int iX = 40000 , iY = 20000;

if( iX + iY > 0)

printf( I am in if );

else

printf( I am in else );}

Tarak


How can i write a macro for finding squreroot of any number?

Aia


main()

{

int x=20,y=35;

x=y++ + x++;

y= ++y + ++x;

printf(%d%d\n,x,y);

}

Answer : 5794

UNDEFINED BEHAVIOR

Thats the answer to it.

Aia


The parts of x = x++ and y = ++y invokes what is called in programming Undefined

Behavior.

When you ch ange the value of a variable more than once in the same sequence point

you are in the land of Undefined Behavior. Meaning you get unpredictable effect.

Depending of compiler implementation or situation. Not a good thing to do.

For further reference read in the C Standards:

ISO Sec. 5.1.2.3, Sec. 6.3, Sec. 6.6, Annex C

Sec. 6.3

and in K&R1 Sec. 2.12 p. 50

K&R2 Sec. 2.12 p. 54

Swapnil


1) what is mean by object?

Ans : Object is a instance of a class.

Mihai Gospodaru


#5 The answer is displayed correctly, because the string begins with a space Cisco

Systems.

It might be useful to display all pieces of code with Courier New font to avoid creating


f 13 30-09-2012 17:19


12/13

confusion among site users.

DIGPAL


hi friends,

somebody can tell me what will be the out put of following code and why?

for (i= printf("a");i


13/13

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Susanta kumar swain


Class afixi {

Function biswa(){

String s= MY Name Is Biswa .;

}

}

i) How to get the value of s out side of the class ?

ii) How to initialize a class in another class ?iii) How to get the method of a class in another class.

iv) Define the class, object, method ?


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C Interview Questions and Answers _ TechInterviews