Agfdhyk

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.

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);

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.)

Because sizeof is not a function, but an operator. It's argument has no side effects.