Can this macro be converted to a function?

Question

While refactoring code and ridding myself of all those #defines that we\'re now taught to hate, I came across this beauty used to calculate the number of elements in a struc

Answer 1

Answered in detail here: Array Size determination Part 1 and here: Array Size determination Part 2.

Answer 2

• function, no template function, yes
• template, I think so (but C++
• templates are not my thing)

Edit: From Doug's code

template <typename T>
uint32_t StructSize()  // This might get inlined to a constant at compile time
{
   return sizeof(T)/sizeof(*T);
}

// or to get it at compile time for shure

class StructSize<typename T>
{
   enum { result = sizeof(T)/sizeof(*T) };
}

I've been told that the 2nd one doesn't work. OTOH something like it should be workable, I just don't use C++ enough to fix it.

A page on C++ (and D) templates for compile time stuff

Answer 3

xtofl has the right answer for finding an array size. No macro or template should be necessary for finding the size of a struct, since sizeof() should do nicely.

I agree the preprocessor is evil, but there are occasions where it is the least evil of the alternatives.

Answer 4

Yes it can be made a template in C++

template <typename T>
size_t getTypeSize()
{
   return sizeof(T)/sizeof(*T);
}

to use:

struct JibbaJabba
{
   int int1;
   float f;
};

int main()
{
    cout << "sizeof JibbaJabba is " << getTypeSize<JibbaJabba>() << std::endl;
    return 0;
}

See BCS's post above or below about a cool way to do this with a class at compile time using some light template metaprogramming.

Answer 5

Windows specific:

There is the macro _countof() supplied by the CRT exactly for this purpose.

A link to the doc at MSDN

Answer 6

Simplfying @KTC's, since we have the size of the array in the template argument:

template<typename T, int SIZE>
int arraySize(const T(&arr)[SIZE])
{
    return SIZE;
}

Disadvantage is you will have a copy of this in your binary for every Typename, Size combination.