Question

Okay so you have and array A[]… that is passed to you in some function say with the following function prototype:

void foo(int A[]);

Okay, as you know it’s kind of hard to find the size of that array without knowing some sort of ending variable or knowing the size already…

Well here is the deal though. I have seem some people figure it out on a challenge problem, and I don’t understand how they did it. I wasn’t able to see their source code of course, that is why I am here asking.

Does anyone know how it would even be remotely possible to find the size of that array?? Maybe something like what the free() function does in C??

What do you think of this??

template<typename E, int size>
int ArrLength(E(&)[size]){return size;}

void main()
{
    int arr[17];
    int sizeofArray = ArrLength(arr);
}

Answer 1

The signature of that function is not that of a function taking an array, but rather a pointer to int. You cannot obtain the size of the array within the function, and will have to pass it as an extra argument to the function.

If you are allowed to change the signature of the function there are different alternatives:

C/C++ (simple):

void f( int *data, int size );             // function
f( array, sizeof array/sizeof array[0] );  // caller code

C++:

template <int N>
void f( int (&array)[N] );                 // Inside f, size N embedded in type
f( array );                                // caller code

C++ (though a dispatch):

template <int N>
void f( int (&array)[N] ) {                // Dispatcher
   f( array, N );
}
void f( int *array, int size );            // Actual function, as per option 1
f( array );                                // Compiler processes the type as per 2