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

Question

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 1

sizeof runs at compile-time, but x++ can only be evaluated at run-time. To solve this, the C++ standard dictates that the operand of sizeof is not evaluated (except for VLAs). The C Standard says:

If the type of the operand [of sizeof] is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

Answer 2

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.

Answer 3

sizeof(foo) tries really hard to discover the size of an expression at compile time:

6.5.3.4:

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.

In short: variable length arrays, run at runtime. (Note: Variable Length Arrays are a specific feature -- not arrays allocated with malloc(3).) Otherwise, only the type of the expression is computed, and that at compile time.

Answer 4

As the operand of sizeof operator is not evaluated, you can do this:

int f(); //no definition, which means we cannot call it

int main(void) {
        printf("%d", sizeof(f()) );  //no linker error
        return 0;
}

Online demo : http://ideone.com/S8e2Y

That is, you don't need define the function f if it is used in sizeof only. This technique is mostly used in C++ template metaprogramming, as even in C++, the operand of sizeof is not evaluated.

Why does this work? It works because the sizeof operator doesn't operate on value, instead it operates on type of the expression. So when you write sizeof(f()), it operates on the type of the expression f(), and which is nothing but the return type of the function f. The return type is always same, no matter what value the function would return if it actually executes.

In C++, you can even this:

struct A
{
  A(); //no definition, which means we cannot create instance!
  int f(); //no definition, which means we cannot call it
};

int main() {
        std::cout << sizeof(A().f())<< std::endl;
        return 0;
}

Yet it looks like, in sizeof, I'm first creating an instance of A, by writing A(), and then calling the function f on the instance, by writing A().f(), but no such thing happens.

Demo : http://ideone.com/egPMi

Here is another topic which explains some other interesting properties of sizeof:

• sizeof taking two arguments

Answer 5

Note

This answer was merged from a duplicate, which explains the late date.

Original

Except for variable length arrays sizeof does not evaluate its arguments. We can see this from the draft C99 standard section 6.5.3.4 The sizeof operator paragraph 2 which says:

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.

A comment(now removed) asked whether something like this would evaluate at run-time:

sizeof( char[x++]  ) ;

and indeed it would, something like this would also work (See them both live):

sizeof( char[func()]  ) ;

since they are both variable length arrays. Although, I don't see much practical use in either one.

Note, variable length arrays are covered in the draft C99 standard section 6.7.5.2 Array declarators paragraph 4:

[...] If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type.

Update

In C11 the answer changes for the VLA case, in certain cases it is unspecified whether the size expression is evaluated or not. From section 6.7.6.2 Array declarators which says:

[...]Where a size expression is part of the operand of a sizeof operator and changing the value of the size expression would not affect the result of the operator, it is unspecified whether or not the size expression is evaluated.

For example in a case like this (see it live):

sizeof( int (*)[x++] )

Answer 6

The execution cannot happen during compilation. So ++i/i++ will not happen. Also sizeof(foo()) will not execute the function but return correct type.