How does “sizeof” work in this helper for determin

Question

I've found this article that brings up the following template and a macro for getting array size:

template<typename Type, size_t Size>
char ( &ArraySizeHelper(Type( &Array )[Size]) )[Size];
#define _countof(Array) sizeof(ArraySizeHelper(Array))

and I find the following part totally unclear. sizeof is applied to a function declaration. I'd expect the result to be "size of function pointer". Why does it obtain "size of return value" instead?

Answer 1

ArraySizeHelper is a function template which returns a char array of size Size. The template takes two parameters, one is type (which is Type), and other is value (which is Size).

So when you pass an object of type, say, A[100] to the function. The compiler deduces both arguments of the template: Type becomes A, and Size becomes 100.

So the instantiated function return type becomes char[100]. Since the argument of sizeof is never evaluated, so the function need not to have definition. sizeof only needs to know the return type of the function which is char[100]. That becomes equivalent to sizeof(char[100]) which returns 100 - the size of the array.

Another interesting point to be noted is that sizeof(char) is not compiler-dependent, unlike other primitive types (other than the variants of char¹). Its ALWAYS 1. So sizeof(char[100]) is guaranteed to be 100.

^{1. Size of all variants of char is ONE, be it char, signed char, unsigned char according to the Standard.}

Answer 2

sizeof is applied to the result of a function call, not a declaration. It therefore gives the size of the return value, which in this case is a reference to an array of chars.

The template causes the array in the return type to have the same number of elements as the argument array, which is fed to the function from the macro.

Finally, sizeof is then applied to a reference to this char array. sizeof on a reference is the same as sizeof on the type itself. Since sizeof(char) == 1, this gives the number of elements in the array.

Answer 3

template<typename Type, size_t Size>
char (&ArraySizeHelper(Type(&Array)[Size]))[Size];
#define _countof(Array) sizeof(ArraySizeHelper(Array))

sizeof is applied to a function declaration. I'd expect the result to be "size of function pointer". Why does it obtain "size of return value" instead?

It's not sizeof ArraySizeHelper (which would be illegal - can't take sizeof a function), nor sizeof &ArraySizeHelper - not even implicitly as implicit conversion from function to pointer-to-function is explicitly disallowed by the Standard, for C++0x see 5.3.3). Rather, it's sizeof ArraySizeHelper(Array) which is equivalent to sizeof the value that the function call returns, i.e. sizeof char[Size] hence Size.