array length using pointers [duplicate]

Question

Array length can be calculated using *(&arr+1)-arr which then simplifies to (&arr)[1]-arr which further simplifies to 1[&arr]-arr.

But when the length is calculated in a function different from where memory allocation has been done, wrong results are computed.

For instance,

#include <iostream> 
#define ARRAY_SIZE(arr) (1[&arr]-arr)      
using namespace std;

void func(int *arr)
{
    cout<<ARRAY_SIZE(arr)<<endl;
}

int main()
{
    int arr[]={1,2,3,4,5};
    cout<<ARRAY_SIZE(arr)<<endl;
    func(arr);
}

This gives the output:

5
8

What accounts for such strange behaviour?

Answer 1

Array length can be calculated using *(&arr+1)-arr

Only if arr is actually an array. Within func, arr is a pointer, so this dereferences a random word of memory to give undefined behavoiur.

There is no way to tell the size of an array given just a pointer to its first element. You could pass the array by reference:

template <size_t N>
void func(int (&arr)[N]) {
    cout<<ARRAY_SIZE(arr)<<endl;
    cout<<N<<endl;               // equivalent, and less weird
}

Using the same technique, we can reimplement ARRAY_SIZE without resorting to the preprocessor or any bizarre pointer arithmetic:

template <size_t N>
size_t ARRAY_SIZE(int (&arr)[N]) {
    return N;
}