int main() { int n = 1; sizeof(n++); printf("%d",n); return 0; } It's part of my code. The output is 1. But why is n not increased by 1?
I tried that for other functions but for others output was 2.
3 Answers
This is because, sizeof is not a function, it is a compile time operator, ans unless the operand is a VLA, the operand is not evaluated at runtime.
Quoting C11, chapter §6.5.3.4
[...] 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.
To add a little on why it does not need to evaluate the operand, from the semantics of the operand,
[....] The size is determined from the type of the operand. [...]
and, the type of the operand is fixed and known at compile time (unless a VLA), so it does not need to evaluate the operand to perform it's job anyway. In your case, since n is of type int
sizeof (n++); is just the same as
sizeof (int); Comments
sizeof is not a function, it's an operator.
sizeof(n++) is equivalent to sizeof n++, which (since n is an int) is equivalent to sizeof (int).
sizeof only looks at the type of its operand; it doesn't evaluate expressions.
(Unless variable-length arrays are involved, but we're going to ignore that for now.)
Comments
Because sizeof is not a function, but an operator. It's argument has no side effects.
1 Comment
Eric Postpischil
Most of the time the operand of
sizeof has no side effects. However, in int n = 3; sizeof(int [n++]); printf("%d\n", n);, n is changed, and “4” is printed.
sizeof n++equates to "sizeof type(n++)" (if this existed). the size of of a type does not change no matter what value it has:sizeof 42==sizeof -3==sizeof isalpha('x')==sizeof (int).