Question

I have this code example, but I don’t understand why changing the values in the array inside
outputUsingArray() are changing the original array.

I would have expected changing the values of the array in outputUsingArray() would only be for a local copy of the array.

Why isn’t that so?

However, this is the behaviour I would like, but I don’t understand why it work.

#include <stdlib.h>
#include <stdio.h>
void outputUsingArray(int array[][4], int n_rows, int n_cols) {
  int i, j;

  printf("Output Using array\n");
  for (i = 0; i < n_rows; i++) {
    for (j = 0; j < n_cols; j++) {
      // Either can be used.
      //printf("%2d ", array[i][j] );
      printf("%2d ", *(*(array+i)+j));
    }
    printf("\n");
  }
  printf("\n");

  array[0][0] = 100;
  array[2][3] = 200;

}

void outputUsingPointer(int (*array)[4], int n_rows, int n_cols) {
  int i, j;

  printf("Output Using Pointer to Array i.e. int (*array)[4]\n");
  for (i = 0; i < n_rows; i++) {
    for (j = 0; j < n_cols; j++) {
      printf("%2d ", *(*(array+i) + j ));
    }
    printf("\n");
  }
  printf("\n");
}

int main() {

  int array[3][4] = { { 0, 1, 2, 3 },
              { 4, 5, 6, 7 },
              { 8, 9, 10, 11 } };

  outputUsingPointer((int (*)[4])array, 3, 4);

  outputUsingArray(array, 3, 4);

  printf("0,0: %i\n", array[0][0]);
  printf("2,3: %i\n", array[2][3]);

  return 0;
}

Answer 1

passing int[] is really passing the pointer to the first element of the array.
passing int* is passing the pointer to the first element of the array as well.

They’re identical.

Since they’re both pointing towards the same part of memory, changing one will change the other one.