To pass an array by reference in C - arrays

Question

I am trying to modify this merge sort codes from GeekstoGeeks with passing an array by reference.
/* C program for Merge Sort */
#include <stdio.h>
#include <stdlib.h>
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
/* create temp arrays */
int L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];
/* Merge the temp arrays back into arr[l..r]*/
i = 0; // Initial index of first subarray
j = 0; // Initial index of second subarray
k = l; // Initial index of merged subarray
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r) {
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l + (r - l) / 2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
/* UTILITY FUNCTIONS */
/* Function to print an array */
void printArray(int A[], int size)
{
int i;
for (i = 0; i < size; i++)
printf("%d ", A[i]);
printf("\n");
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 12, 11, 13, 5, 6, 7 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printf("Given array is \n");
printArray(arr, arr_size);
mergeSort(arr, 0, arr_size - 1);
printf("\nSorted array is \n");
printArray(arr, arr_size);
return 0;
}
To something like this:
void mergeSort(int* arr[], int l, int r);
void merge(int* arr[], int l, int m, int r);
I thought it is a simple task because I could simply change every arr to *arr in the functions to dereference them. The program is still runnable but gives me a bunch of unknown numbers.
Then I searched the stack overflow until I see this post:
https://stackoverflow.com/questions/49751409/swapping-2-arrays-in-c#:~:text=Since%20pointer%20is%2064%20bit,two%20int%20which%20get%20swapped.&text=In%20C%20an%20array%20is,they%20are%20pointers%20to%20arrays
It also passes an array by reference. (int* a and int* b)
void swapArray( int *a, int *b, size_t n )
{
for ( size_t i = 0; i < n; i++ )
{
int tmp = a[i];
a[i] = b[i];
b[i] = tmp;
}
}
But in reverse, when I change it to
void swapArray( int a, int b, size_t n )
{
for ( size_t i = 0; i < n; i++ )
{
int tmp = a[i];
a[i] = b[i];
b[i] = tmp;
}
}
It doesn't work, again. Could someone explain to me why and how can I modify my merge sort codes properly? Or where I should look for? I am so confused about this. Any help is appreciated.

Answer 1

int *arr[] is an array of pointers. Or rather it's a pointer to a pointer (it's really int **arr).
To pass a pointer to an array (which almost never is needed) you need e.g. int (*arr) [6]. And yes, the size of the array is then mandatory (variable-length arrays are valid).
But as I said, passning pointers to arrays is almost never needed, because arrays naturally decays to pointers (to their first element).
And as an argument declaration int arr[] is equal to int *arr.
That is, with
void merge(int arr[], int l, int m, int r)
you already pass the array by "reference".

Answer 2

The subscript operator, [ ], has higher precedence than the dereference operator, *. So int *arr[] declares arr to be an array of pointers to int. To make it a pointer to an array, you need int (*arr)[].
Similarly, in expressions, *arr[3] attempts to take element 3 of an array arr and dereference it. Instead you need (*arr)[3] to use *arr to get the array and then [3] to get element 3.