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main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
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Answer / odelu vanga
*f2=*f2+(*f=*f2+(a=a+2.5))
a=a+2.5=4
so *f2=4
*f2=*f2+4=8
now *f2=8
so *f2=*f2+8=16
ans:- 16 16 16
| Is This Answer Correct ? | 18 Yes | 3 No |
Answer / susie
Answer :
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So
changes through f1 and f2 ultimately affects only the value
of a.
| Is This Answer Correct ? | 6 Yes | 3 No |
Answer / kamlesh meghwal
LOL@Govind Verma....how u r gettin 888..hehe..!!1
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / forgot_my_name
Since we are modifying same variable three times in the same line (before sequence point ) so we broke the rules, so whatever compiler says would be right..
@Kamlesh Meghwal : In GCC compiler ans is 8 8 8.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / vamshikrishna
12 12 12
explanition.
printf takes the last updated value and prints for every var
i.e,
a=a+2.5=4
so f2=4
f2=f2+4=8
here again f2 =4 "since f2=&a"
so f2=f2+4=12
| Is This Answer Correct ? | 3 Yes | 3 No |
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C Code 
