Question

Is the return value of this function well defined by the C standard?

int foo()
{
    char buff[128];
    // This is the important line:
    if ((void*)buff == (void*)(&buff))
        return 1;
    else
        return 0;
}

What is the result of foo? On gcc and clang, it will always be 1, but I do not think that this is guaranteed by the standard.

Answer 1

It will always return 1.

An array is a contiguously allocated non-empty set of objects with the element type – arrays are not allowed to have padding before or between elements. It follows that a pointer to an array points at the same location as a pointer to the first element, although it has a different type.

&buff is the address of the variable buff. The variable buff in this case is an array, so &buff is the address of an array – and just like the address of a struct variable is the same location as the address of its first member (although a different type), the address of an array variable is the same location as the address of its first member.

§6.5.9 tells us:

6 Two pointers compare equal if and only if both are null pointers,
both are pointers to the same object (including a pointer to an object
and a subobject at its beginning)…

In this case the “pointer to an object” is the pointer to the array, and the “pointer to the subobject at its beginning” is the pointer to the first element of the array.

In other words, it returns 1 for much the same reason that this code does:

int foo()
{
    struct { int a; int b; } s;

    return (void*)&s == (void*)&s.a;
}