Trouble passing a 2d array to function after malloc - c

Question

I've gotten back into C recently after a long absence, and I can't for the life of me remember how to pass a 2D matrix to a function and use it after mallocating it. I believe I've passed the pointer correctly, but I can't seem to actually access anything in the matrix and can't figure out why.
This is what I have written:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <stdbool.h>
void matrixOperation(int *arr, int m, int n)
{
printf("in matrixOperation\n ");
int i, j;
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
printf("%d\n", *((arr+i*n) + j));
}
int main()
{
int i,j,count;
int row, col;
//------------------------------
printf("Number of rows?\n ");
scanf("%d", &row);
printf("Number of columns?\n ");
scanf("%d", &col);
//------------------------------
int* arr[row];
for (i = 0; i < row; i++)
arr[i] = (int*)malloc(col * sizeof(int));
count = 0;
for (i = 0; i < row; i++)
for (j = 0; j < col; j++)
arr[i][j] = ++count;
for (i = 0; i < row; i++)
for (j = 0; j < col; j++)
printf("%d\n",arr[i][j]);
// We can also use "print(&arr[0][0], m, n);"
matrixOperation((int *)arr, row, col);
for (int i = 0; i < row; i++)
free(arr[i]);
return 0;
}
The goal was to have it accept user input for the size of the matrix, fill each index with a count and then pass it to a function which would print it out. However when I try it the print statement just outputs random numbers which I assume are what the pointers are pointing at.
I would like matrixOperation to print out the same thing as the print statement in main and I can't quite figure out what I've done wrong.

Answer 1

Change the function signature and you should be able to use the same piece of code.
void matrixOperation(int **arr, int m, int n)
{
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
printf("%d\n",arr[i][j]);
}
Call with:
matrixOperation(arr, row, col);
Remember when an array is passed to a function, it "decays" into a pointer. You have an array of pointers. That decays to a pointer to pointers.

Answer 2

You declared an array of pointers:
int* arr[row];
So pass this array to the function.
The function will be declared like:
void matrixOperation(int **arr, int m, int n)
{
printf("in matrixOperation\n ");
int i, j;
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
printf("%d\n", *( *( arr + i ) + j ) );
}
and call the function like:
matrixOperation( arr, row, col );
Though it will be better to define the function like:
void matrixOperation(int **arr, int m, int n)
{
printf("in matrixOperation\n ");
for ( int i = 0; i < m; i++ )
{
for ( int j = 0; j < n; j++ )
{
printf("%d ", *( *( arr + i ) + j ) );
}
putchar( '\n' );
}
}
Pay attention to that this expression with the pointer arithmetic:
*( *( arr + i ) + j )
can be rewritten using the subscript operator the following way:
arr[i][j]
The both expressions are equivalent each other.
As for your call of the function with casting the array designator:
matrixOperation((int *)arr, row, col);
then it is wrong because you do not have one extent of memory. You allocated several extents of memory.