sizeof operator doesn't evaluate the subexpression?

Question

Possible Duplicate:
Why does sizeof(x++) not increment x?

#include<stdio.h>
int main(void)
{
  double num=5.2;
  int var=5;
  printf("%d\t",sizeof(!num));
  printf("%d\t",sizeof(var=15/2));
  printf("%d",var);
  return 0;
}

The program gave an output 4 4 5. I didn't quite get why this happened.

Why was the first output 4?
Why didn't the value of var get updated to 7?

How does the sizeof operator work?

while i accept my question happened to be a duplicate of another question on SO, a more appropriate heading to the other question and I might have never put up the question. Didn't turn up in search. Anyways thanks guys. — sasidhar, Jul 24 '12 at 11:58

score 3 · Accepted Answer · edited May 23 '17 at 11:55

3

The expressions

!num
var=15/2

both evaluate to an int. sizeof(int) is always 4 on your platform.

As for why var is not updated:

sizeof is a compile-time operator, so at the time of compilation sizeof and its operand get replaced by the result value. The operand is not evaluated (except when it is a variable length array) at all; only the type of the result matters.

https://stackoverflow.com/a/8225813/141172

edited May 23 '17 at 11:55

Community

1
1

answered Jul 24 '12 at 05:34

Eric J.

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how come `var` is not updated?? – sasidhar Jul 24 '12 at 05:36
@sasidhar: Look at the comment by Mystical under your question. – nhahtdh Jul 24 '12 at 05:37

steffen · Answer 2 · 2012-07-24T12:10:36.050

1

The sizeof operator shows you the size of the darts type of its argument. It does not return the result of the argument. On your platform an integer is 4 bytes long, that's what you learn here. If you want the result, just cut out the sizeof.

edited Jul 24 '12 at 12:10

answered Jul 24 '12 at 05:39

steffen

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sizeof operator doesn't evaluate the subexpression?

2 Answers2