C pointer to array well-defined behavior

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

Are they [buff and &buff] guaranteed to have the same value or is that something that just happens with gcc and clang? I ask because I don't think &buff is "the address of the start of the array," but the address of the variable buff.

They are guaranteed to be the same value. In C arrays can be considered "special" in the sense that they're not pointers and the only time they're treated as such is when they're passed to a function and they decay to pointers.

For that reason:

buff == &buff && buff == &buff[0]

However buff and &buff are of two different types, where &buff is a pointer to the array (they still refer to the same spot!). You can use sizeof on the two to see this difference for yourself.

That is not the same for a pointer, which as you say will return the address of that actual pointer variable. An array also isn't a pointer to itself since in order for a pointer to point to itself, the address of that pointer would need to be stored in that pointer variable.

An array name is simply a label used by the compiler that corresponds to a contiguous block of memory. It has special properties, the above being one of them.