#include<stdio.h> main(0 { printf("\n %d %d %d",sizeof(3),si

#include<stdio.h>
main(0
{
printf("\n %d %d %d",sizeof(3),sizeof("3"),sizeof(3));
}

Question Posted / sandhyait

4 Answers
10266 Views
HCL, I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

#include<stdio.h> main(0 { printf("\n %d %d %d",sizeof(3),sizeof("3"),..

Answer / nithya

the above code output is

syntax error

this code will be change from

#include<stdio.h>
main()
{
printf("\n %d %d %d",sizeof(3),sizeof("3"),sizeof(3));
}

the out put is

222

Is This Answer Correct ?

9 Yes

1 No

#include<stdio.h> main(0 { printf("\n %d %d %d",sizeof(3),sizeof("3"),..

Answer / a.c.pattanaik

code is error due to declaration

if code is this
main()
{
printf("\n %d %d %d",sizeof(2),sizeof("a"),sizeof(2));
}

Ans-4 2 4

Is This Answer Correct ?

9 Yes

5 No

#include<stdio.h> main(0 { printf("\n %d %d %d",sizeof(3),sizeof("3"),..

Answer / ajay

syntax error

Is This Answer Correct ?

1 Yes

1 No

#include<stdio.h> main(0 { printf("\n %d %d %d",sizeof(3),sizeof("3"),..

Answer / abhishek kumar verma

you have made a typing mistake if code will be this

#include<stdio.h>
main()
{
printf("\n %d %d %d",sizeof(3),sizeof("3"),sizeof(3));
}

then output will be 4 2 4
because in first it will print the size of an integer and in second it will print the size of an string and in third it will print the size of integer again .

Is This Answer Correct ?

4 Yes

5 No

Post New Answer