Why does sizeof(x++) not increment x?

c sizeof

Neigyl R. Noval · Nov 22, 2011 · Viewed 30.5k times · Source

Here is the code compiled in dev c++ windows:

#include <stdio.h>

int main() {
    int x = 5;
    printf("%d and ", sizeof(x++)); // note 1
    printf("%d\n", x); // note 2
    return 0;
}

I expect x to be 6 after executing note 1. However, the output is:

4 and 5

Can anyone explain why x does not increment after note 1?

Answer

From the C99 Standard (the emphasis is mine)

6.5.3.4/2

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

Why does sizeof(x++) not increment x?

Answer

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