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Can someone explain this template code that gives me the size of an array?

template <typename T,unsigned S>
unsigned ArraySize(const T (&v)[S])
{
    return S;
}

I understand the T and S, but my question is why do we have to declare v as a reference variable? Or maybe I’m misunderstanding this whole thing.

I appreciate the help!

Answer 1

The function accepts an array by reference, and because of this, type of element T and size S is deduced by the compiler. So it returns S which is nothing but the size of the array. In the absence of reference, it would decay into a pointer type. So there is no difference between these:

void f(int v[100]);  //declaration of f
void f(int v[200]);  //redeclaration of f
void f(int v[]);     //redeclaration of f
void f(int *v);      //redeclaration of f

All are exactly same. You can pass array of any size to all of these functions.

---

Coming back to ArraySize, the returned value of this function cannot be used as constant expression:

int a[10];
SomeClassTemplate<ArraySize(a)> obj; //error

See error : http://ideone.com/4mdJE

So a better implementation would be this:

template <typename T,unsigned S>
char (&ArraySizeHelper(const T (&v)[S]))[S];  //no need to define it!
#define ArraySize(a) sizeof(ArraySizeHelper(a))

Now this is perfectly fine:

int a[10];
SomeClassTemplate<ArraySize(a)> obj; //ok

See ok : http://ideone.com/Zt3UY