Passing a pointer by reference

Question

#include <iostream>

using namespace std;

void merge(int *& toMerge, int lo, int mid, int hi)
{
    int merged[hi+1];
    int i = lo, j = mid+1;
    for (int k = lo; k <= hi; k++)
    {
        if (i > mid) {merged[k] = toMerge[j]; j++;}
        else if (j > hi) {merged[k] = toMerge[i]; i++;}
        else if (toMerge[j] < toMerge[i]) {merged[k] = toMerge[j]; j++;}
        else {merged[k] = toMerge[i]; i++;}
    }
    toMerge = merged;
}

int main(int argc, const char * argv[])
{

    int x[8] = {1,7,4,6,2,7,3,2};
    merge(x, 0, 7, 3);
    return 0;
}

I am trying to pass a pointer by reference here such that the end result will be that the array x will be sorted. However, my compiler is throwing an error that there is no matching function call for the merge(x, 0, 7, 3).

I am not sure as to why this is happening because the merge function requires a pointer, and x is a pointer to the first element in the array -- so it should work. Why doesn't it?

Answer 1

An array decays to a pointer when you pass it as an argument to a function call but not to a reference to a pointer.

Imagine this:

void foo(int*& p)
{
   p = new int;
}

int main()
{
   int arr[6] = {0};

    foo(arr);
   // You don't want arr to be an invalid object here.
   // If the compiler allowed you to do that, that's what will happen.
   // Not only that, foo(arr); is equivalent to:
   // arr = new int;
   // which is not allowed in the language.
}

Answer 2

You really should NOT pass a reference, because you don't really want merge changing the value of the pointer. You just want to allow it to change the values stored in the array, which it can do without the reference (i.e., just from the pointer).