C++ Pointers with 2D Arrays

Question

How do you you implement function that takes a pointer int* to a 2D array as an input? My current code is:

#include <iostream>
using namespace std;

int main (void){
  int M [4][4] = {
    {1,2,3,4},
    {5,6,7,8},
    {9,10,11,12},
    {13,14,15,16},
  };
  int* Mat = M;
  myFunc(Mat);
}

void myFunc(int* Matrix)

check this : https://stackoverflow.com/questions/8617466/a-pointer-to-2d-array — ajaysinghnegi, Feb 06 '19 at 02:07
I, too, cannot tell what you are asking. A question typically contains a question mark. — Richard, Feb 06 '19 at 02:13
"I am currently working on a C++ function that I am working on" is probably this month's best. — bipll, Feb 06 '19 at 02:42
You must have been given some description of what the argument is supposed to be. (My guess is that you're expected to store your data in a one-dimensional array and compute the one-dimensional index from the two-dimensional ones.) — molbdnilo, Feb 06 '19 at 03:25
You can't pass `M` as `int*`, you can pass a row of `M`, but not the entire `M`. — David C. Rankin, Feb 06 '19 at 05:37

Brady Dean · Answer 1 · 2019-02-06T03:01:10.227

1

MxN arrays decay to a pointer to the first row (length N). If you want a pointer to the beginning then you need to allow the first row to decay to a pointer to the first element. Also note what @Pete Becker says below.

#include <iostream>

void myFunc(int* Matrix);

int main (void){
  int M [4][4] = {
    {1,2,3,4},
    {5,6,7,8},
    {9,10,11,12},
    {13,14,15,16},
  };

  int* Mat = M[0];
  myFunc(Mat);
}

edited Feb 06 '19 at 03:01

answered Feb 06 '19 at 02:08

Brady Dean

3,107
4
24
45

**Two** dimensional arrays decay to a pointer to the first row. **Thee** dime signal arrays decay to a pointer to the first two dimensional array. Etc. – Pete Becker Feb 06 '19 at 02:54

score 0 · Answer 2 · answered Feb 06 '19 at 05:52

The answer was quite simple and it only took me like 4 hours (gotta love coding). So take out int* Mat = M[0]; and when calling the function simply recast as (int*)

#include <iostream>

void myFunc(int* Matrix);

int main (void){
  int M [4][4] = {
    {1,2,3,4},
    {5,6,7,8},
    {9,10,11,12},
    {13,14,15,16},
  };

  myFunc((int*)M);
}

May · Answer 3 · 2019-02-06T05:32:28.887

-2

M pointing to the first row, but its value is the same as pointer to the first element which is M[0][0], so you might change the interpretation for compiler:

int* Mat = reinterpret_cast<int*>(M);

edited Feb 06 '19 at 05:32

answered Feb 06 '19 at 02:49

May

99
7

2

The type of `M` is `int[4][4]` – M.M Feb 06 '19 at 03:03
that's not true, do you define a variable like this `int p[4][4] = M;`? this won't even compile. – May Feb 06 '19 at 03:23
It is true, the C standard says so. The reason `int p[4][4] = M;` does not compile is that array initializers must be a braced list of initializers for each element – M.M Feb 06 '19 at 05:06

C++ Pointers with 2D Arrays

3 Answers3