c++ dynamic array initialization with declaration

Question

I have function like this:

void findScarf1(bool ** matrix, int m, int n, int radius, int connectivity); 

and in main function I create 2d dynamic array to pass in this function

    bool matrix[6][7] = {
    {0, 0, 1, 1, 1, 0, 0},
    {0, 0, 1, 1, 1, 0, 0},
    {0, 0, 1, 1, 1, 0, 0},
    {0, 0, 1, 1, 1, 0, 0},
    {0, 0, 1, 1, 1, 0, 0},
    {0, 0, 1, 1, 1, 0, 0}
};

The problem is:

findScarf1(matrix, 6, 7, 3, 4);

causes error C2664: 'findScarf1' : cannot convert parameter 1 from 'bool [6][7]' to 'bool **'

How to initialize array compactly(simultaneously with declaration)?

p.s. sorry if it's duplicate question but I've spent 1.5 hours figuring it out

Answer 1

Technically, a 2D array is an array of 1D arrays. So it cannot convert into pointer to pointer. It can convert into pointer to array, though.

So this should work:

void findScarf1(bool (*matrix)[7], int m, int n, int radius, int connectivity); 

Here bool (*matrix)[7] declares a pointer to array of 7 bool.

Hope that helps.

Answer 2

If you look at how your array is laid out in memory, and compare it how a pointer-to-pointer "matrix" is laid out, you will understand why you can't pass the matrix as a pointer to pointer.

You matrix is like this:

[ matrix[0][0] | matrix[0][1] | ... | matrix[0][6] | matrix[1][0] | matrix[1][1] | ... ]

A matrix in pointer-to-pointer is like this:

[ matrix[0] | matrix[1] | ... ]
  |           |
  |           v
  |           [ matrix[1][0] | matrix[1][1] | ... ]
  v
  [ matrix[0][0] | matrix[0][1] | ... ]

You can solve this by changing the function argument:

bool (*matrix)[7]

That makes the argument matrix a pointer to an array, which will work.

---

And by the way, the matrix variable you have is not dynamic, it's fully declared and initialized by the compiler, there's nothing dynamic about it.

Answer 3

You may define the function as:

void findScarf1(bool * matrix, int m, int n, int radius, int connectivity);

Or

void findScarf1(bool matrix[][7], int m, int n, int radius, int connectivity);

No matter how many dimensions an array has, it's just a block of linear memory.

When you use the first manner, you may need to do a cast when call this function:

findScarf1((bool *)marix, 6, 7, 3, 4);

Answer 4

My Following example may be helpful for you:

#include<stdio.h>
void f(int (*m)[7]){  // As Nawaz answred
 printf("%d\n", m[3][3]);

}
void _f(int m[6][7]){ // As I commented to your question
 printf("%d\n", m[3][3]);

}
void _f_(int m[][7]){// Second form of Nawaz's answe 
 printf("%d\n", m[3][3]);

}
void f_(int (*m)[6][7]){// Pointer of 2D array
 printf("%d\n", (*m)[3][3]);
}

int main(){

    int matrix[6][7] = {
    {0, 0, 1, 1, 1, 0, 0},
    {0, 0, 1, 3, 1, 0, 0},
    {0, 0, 1, 4, 1, 0, 0},
    {0, 0, 1, 5, 1, 0, 0},
    {0, 0, 1, 6, 1, 0, 0},
    {0, 0, 1, 7, 1, 0, 0}
    };
    f(matrix);
    _f(matrix);
    _f_(matrix);    
    f_(&matrix);
    return 1;
}    

question not tanged to c, but I compiled with gcc (I have not installed g++).

~$ gcc  -Wall -pedantic 2d.c
~$ ./a.out 
5
5
5
5

_{I was not intended to post an answer, but because I commented wrong to Nawaz answer so during an experiment I written this code.}

Here one can find it working at codepacde