c++ - Calculating size of an array

Question

I am using the following macro for calculating size of an array:

#define G_N_ELEMENTS(arr) ((sizeof(arr))/(sizeof(arr[0])))  

However I see a discrepancy in the value computed by it when I evaluate the size of an array in a function (incorrect value computed) as opposed to where the function is called (correct value computed). Code + output below. Any thoughts, suggestions, tips et al. welcome.

DP

#include <stdio.h>

#define G_N_ELEMENTS(arr) ((sizeof(arr))/(sizeof(arr[0])))

void foo(int * arr) // Also tried foo(int arr[]), foo(int * & arr) 
                    // - neither of which worked
{
   printf("arr : %x
", arr);
   printf ("sizeof arr: %d
", G_N_ELEMENTS(arr));
}

int main()
{
   int arr[] = {1, 2, 3, 4};

   printf("arr : %x
", arr);
   printf ("sizeof arr: %d
", G_N_ELEMENTS(arr));

   foo(arr);
}

Output:

arr : bffffa40
sizeof arr: 4
arr : bffffa40
sizeof arr: 1

Answer 1

That's because the size of an int * is the size of an int pointer (4 or 8 bytes on modern platforms that I use but it depends entirely on the platform). The sizeof is calculated at compile time, not run time, so even sizeof (arr[]) won't help because you may call the foo() function at runtime with many different-sized arrays.

The size of an int array is the size of an int array.

This is one of the tricky bits in C/C++ - the use of arrays and pointers are not always identical. Arrays will, under a great many circumstances, decay to a pointer to the first element of that array.

There are at least two solutions, compatible with both C and C++:

• pass the length in with the array (not that useful if the intent of the function is to actually work out the array size).
• pass a sentinel value marking the end of the data, e.g., {1,2,3,4,-1}.