Calculate Length of Array in C by Using Function

Question

I want to make a FUNCTION which calculates size of passed array.

I will pass an Array as input and it should return its length. I want a Function

int ArraySize(int * Array   /* Or int Array[] */)
{
   /* Calculate Length of Array and Return it */

}

void main()
{
  int MyArray[8]={1,2,3,0,5};
  int length;

  length=ArraySize(MyArray);

  printf("Size of Array: %d",length);

}

Length should be 5 as it contains 5 elements though it's size is 8 (Even 8 will do but 5 would be excellent)

I tried this:

int ArraySize(int * Array)
{

  return (sizeof(Array)/sizeof(int));

}

This won't work as "sizeof(Array)" will retun size of Int Pointer. This "sizeof" thing works only if you are in same function.

Actually I am back to C after lots of days from C# So I can't remember (and Missing Array.Length())

Regards!

Answer 1

You cannot calculate the size of an array when all you've got is a pointer.

The only way to make this "function-like" is to define a macro:

#define ARRAY_SIZE( array ) ( sizeof( array ) / sizeof( array[0] ) )

This comes with all the usual caveats of macros, of course.

Edit: (The comments below really belong into the answer...)

1. You cannot determine the number of elements initialized within an array, unless you initialize all elements to an "invalid" value first and doing the counting of "valid" values manually. If your array has been defined as having 8 elements, for the compiler it has 8 elements, no matter whether you initialized only 5 of them.
2. You cannot determine the size of an array within a function to which that array has been passed as parameter. Not directly, not through a macro, not in any way. You can only determine the size of an array in the scope it has been declared in.

The impossibility of determining the size of the array in a called function can be understood once you realize that sizeof() is a compile-time operator. It might look like a run-time function call, but it isn't: The compiler determines the size of the operands, and inserts them as constants.

In the scope the array is declared, the compiler has the information that it is actually an array, and how many elements it has.

In a function to which the array is passed, all the compiler sees is a pointer. (Consider that the function might be called with many different arrays, and remember that sizeof() is a compile-time operator.

You can switch to C++ and use <vector>. You can define a struct vector plus functions handling that, but it's not really comfortable:

#include <stdlib.h>

typedef struct
{
    int *  _data;
    size_t _size;
} int_vector;

int_vector * create_int_vector( size_t size )
{
    int_vector * _vec = malloc( sizeof( int_vector ) );
    if ( _vec != NULL )
    {
        _vec._size = size;
        _vec._data = (int *)malloc( size * sizeof( int ) );
    }
    return _vec;
}

void destroy_int_vector( int_vector * _vec )
{
    free( _vec->_data );
    free( _vec );
}

int main()
{
    int_vector * myVector = create_int_vector( 8 );
    if ( myVector != NULL && myVector->_data != NULL )
    {
        myVector->_data[0] = ...;
        destroy_int_vector( myVector );
    }
    else if ( myVector != NULL )
    {
        free( myVector );
    }
    return 0;
}

Bottom line: C arrays are limited. You cannot calculate their length in a sub-function, period. You have to code your way around that limitation, or use a different language (like C++).