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

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.