Question

I am trying to learn C++ function templates.I am passing an array as pointer to my function template. In that, I am trying to find the size of an array. Here is the function template that I use.

template<typename T>
T* average( T *arr)
{
    T *ansPtr,ans,sum = 0.0;    

    size_t sz = sizeof(arr)/sizeof(arr[0]);
    cout<<"\nSz is "<<sz<<endl;

    for(int i = 0;i < sz; i++)
    {
        sum = sum + arr[i];
    }
    ans = (sum/sz);
    ansPtr = &ans;
    return ansPtr;
}

The cout statement displays the size of arr as 1 even when I am passing the pointer to an array of 5 integers. Now I know this might be a possible duplicate of questions to which I referred earlier but I need a better explanation on this.

Only thing I could come up with is that since templates are invoked at runtime,and sizeof is a compile time operator, compiler just ignores the line

   int sz = sizeof(arr)/sizeof(arr[0]);

since it does not know the exact type of arr until it actually invokes the function.
Is it correct or am I missing something over here? Also is it reliable to send pointer to an array to the function templates?

Answer 1

 T *arr

This is C++ for “arr is a pointer to T“. sizeof(arr) obviously means “size of the pointer arr“, not “size of the array arr“, for obvious reasons. That’s the crucial flaw in that plan.

To get the size of an array, the function needs to operate on arrays, obviously not on pointers. As everyone knows (right?) arrays are not pointers.

Furthermore, an average function should return an average value. But T* is a “pointer to T“. An average function should not return a pointer to a value. That is not a value.

Having a pointer return type is not the last offense: returning a pointer to a local variable is the worst of all. Why would you want to steal hotel room keys?

template<typename T, std::size_t sz>
T average( T(&arr)[sz])
{
    T ans,sum = 0.0;    

    cout<<"\nSz is "<<sz<<endl;

    for(int i = 0;i < sz; i++)
    {
        sum = sum + arr[i];
    }
    ans = (sum/sz);
    return ans;
}