Warning: Expected ‘int **’ but argument is of type ‘int (*)[(sizetype)(n)]’ [duplicate]

Question

Following code:

#include <stdio.h>

void printSpiral(int **M, int row1, int row2, int col1, int col2) {
  if (!((row1 <= row2) && (col1 <= col2)))
    return;
  int i;
  for (i = col1; i <= col2; i++)
    printf("%d ", M[row1][i]);
  for (i = row1; i <= row2; i++)
    printf("%d ", M[i][col2]);
  for (i = col2; i >= col1; i--)
    printf("%d ",M[row2][i]);
  for (i = row2; i >= row1; i--)
    printf("%d ",M[i][col1]);
  printSpiral(M,row1+1,row2-2,col1+1,col2-1);
}

int main() {
  int n;
  scanf("%d",&n);
  int M[n][n];
  int i, j;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      scanf("%d",&M[i][j]);
  printSpiral(M,0,n-1,0,n-1);
  return 0;
}

gives following warning:

spiral.c: In function ‘main’:

spiral.c:26:3: warning: passing argument 1 of ‘printSpiral’ from incompatible pointer     
type [enabled by default]
printSpiral(M,0,n-1,0,n-1);`

spiral.c:3:6: note: `expected ‘int **’ but argument is of type ‘int (*)[(sizetype)(n)]’`
void printSpiral(int **M, int row1, int row2, int col1, int col2) {

I have seen this warning for the first time. What does it mean?

Answer 1

There are two problems with this code. First you are passing argument of int (*)[(sizetype)(n)] (pointer to an array of n integers, this is the type your 2D array M decays to when you pass it to a function) to your function which is expecting an argument of int **. Always remember Arrays are not pointers. One possible solution is you can change your function's first parameter to int (*)[(sizetype)(n)] type.

void printSpiral(int (*M)[n], int row1, int row2, int col1, int col2) {  

But by doing this the second problem will comes into light and that is because of you declared M as variable length array and n is not known to the function. This issue can be resolved by passing n from main to your function and hence change your function definition to

void printSpiral(int n, int (*M)[n], int row1, int row2, int col1, int col2) {

Answer 2

The compiler wants to tell you that the functions expects a int **, but you provide it an argument of type int (*)[(sizetype)(n)].

The first is a pointer to (the first element of an array of) int *, the second one is a pointer to an array/arrays of given length.

Compare what both means in memory and why this cannot work this way.

The solution would be to provide the function a pointer to one element an at least the total number of columns. (If you want the function to do bounds-checking, it needs the number of rows as well.) Then you have to calculate the indexes manually.