How do I add two matrixes using pointers in c?

Question

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

void adder(int **matrix1, int **matrix2, int **matrix3) {

  for(int i = 0; i < 3; i++){
    for(int j = 0; j < 3; j++){
      matrix3[i][j] = matrix1[i][j] + matrix2[i][j];
    }
  }

}

int main() {

int matrix1[3][3] = {{1,1,1},{1,1,1},{1,1,1}};
int matrix2[3][3] = {{2,2,2},{2,2,2},{2,2,2}};
int matrix3[3][3];

adder(matrix1, matrix2, matrix3);

for(int i = 0; i < 3; i++){
  for(int j = 0; j < 3; j++){
    printf("%d\t", matrix3[i][j]);
  }
  printf("\n");
}
}

When doing this there are no errors but warning saying "passing argument from incompatible pointer". I've tried working with the adder inputs as pointers using int *matrix1 but this doesn't work either.

What can I do?

replace int **matrix1 with int matrix1[3][3], do the same for the other arguments. Or allocate the matrix1 using malloc — dreamcrash, Nov 22 '16 at 19:08
`void adder(int (*matrix1)[3], int (*matrix2)[3], int (*matrix3)[3]) {` — BLUEPIXY, Nov 22 '16 at 19:09
Maybe [question#5329107](http://stackoverflow.com/questions/5329107/passing-a-pointer-representing-a-2d-array-to-a-function-in-c) might help — amanshri93, Nov 22 '16 at 19:10
Yeah this works but the assignment is to specifically use pointers. — Ryan Hitt, Nov 22 '16 at 19:12
@RyanHitt the point of above comments could be that your function has no idea of the width of each row of the 2D arrays. All it gets is pointers, which are only meaningful in context. — Weather Vane, Nov 22 '16 at 19:16
There are no matrices (aka 2D arrays) in your function arguments. See @BLUEPIXY comment. A pointer is not an array! — too honest for this site, Nov 22 '16 at 19:23
Possible duplicate of [How to pass 2d array to a function?](http://stackoverflow.com/questions/38408760/how-to-pass-2d-array-to-a-function) — too honest for this site, Nov 22 '16 at 19:24

score 1 · Answer 1 · answered Nov 22 '16 at 19:16

int ** is not the same type as int[3][3].

int ** is (as used here) a pointer to an array of int * pointers, where each of those pointers addresses an array of ints. In graphical form:

            +-------+     +-----+
int **a --> | int * ----> | int |
            | int * |     | int |
            | int * |     | int |
            +-------+     +-----+

int[3][3] is a single array of 9 ints, which is intended to be accessed as a 3x3 array. A variable of this type can be treated as an int *, but the array that points to behaves as a single pointer, not a double pointer. It doesn't have the extra level of indirection seen in int **.

score 0 · Answer 2 · answered Nov 22 '16 at 19:22

0

Even an array is type-compatible with a pointer, this does not work for multi-dimensional arrays. You only can convert the outermost dimension. To make a long story short:

This would lose the information about size. For the outermost this is okay, because it always have a linear layout. But the inner arrays would lose its size.
A pointer to pointer to an object needs at least two pointers. This is obvious. A mutli-dimensional array does not have additional pointer in its inside.

There are some solutions:

Make the type of the parameter

a pointer to the inner array (dimensions-1): int (*m)[3]
an array, whose outer dimension is incomplete: int m[][3]
the real dimension: int m[3][3]
or – in your case – make it a simple pointer to the int and iterate over an 1-dimensional array, what will work in your case: int *m

answered Nov 22 '16 at 19:22

Amin Negm-Awad

16,582
3
35
50

Arrays are **not** type-compatible with a pointer! Try `sizeof(int[9]) == sizeof(int *)`. And as a multidimensional array is just a 1D array of an array, there is no exception for those. – too honest for this site Nov 22 '16 at 19:25
I know that. This is, way I called it type-**compatible**, not equivalent. Compatibility and equivalence are not the same. – Amin Negm-Awad Nov 22 '16 at 19:32
They are definitively also not type-compatible (if they were, they' s also be equivalent!). That's **exactly** the reason why `int [][]` cannot be converted to `int **`. There is already enough confusion by beginners and such wrong statements just make it worse. – too honest for this site Nov 22 '16 at 20:35
No, that's not the reason. The reason, that [][] cannot be converted to ** and vice versa is, that the objects are different as explained in my answer. Maybe you misunderstand the term *compatible*. Compatibility and equivalence are still different things. This is the reason, why there are two different terms for it. – Amin Negm-Awad Nov 23 '16 at 04:32
The term _compatible types_ is exactly defined in the standard! I don't missunderstand it. Don't try to enforce your very own definition of such well-defined terms! – too honest for this site Nov 23 '16 at 11:46

How do I add two matrixes using pointers in c?

2 Answers2