How to modify dynamic matrix?

Question

I’m having trouble modifying a dynamic matrix already created. The following example depicts my problem. In this example the program compiles but gives error of execution ( nothing appears and shows the dialog saying that the program stopped working).

#include <iostream>
#include <stdio.h>
#include <stdlib.h>

using namespace std;

void Alocar(int **m,int rows,int cols){

    m = (int **)malloc((rows)*sizeof(int*));

    for(int i=0;i<rows;i++){
         m[i] = (int *)malloc(cols*sizeof(int));   
    }

}

void Modificar(int **m,int rows,int cols){


     for(int i=0; i<rows ;i++)
     {
        for(int j=0; j<cols ; j++)
        {

          m[i][j]= 0;
        }
     }
     cout << " Matriz zerada" << endl;
      for(int i=0; i<rows ;i++)
     {
        for(int j=0; j<cols ; j++)
        {

          cout << m[i][j]<< " ";
        }
        cout << endl;
     }

}

int main()
{
     int **M;

     Alocar(M,100,100);

     Modificar(M,100,100);

    return 0;

}

Does anyone know why this kind of problem is occurring?

Answer 1

The problem in the code shown comes in the function Alocar. When does m = ... within the function Alocar is changing a copy of the pointer in the main what makes the main continue without being initialized.

It can solve this problem in some ways:

• Returning the new pointer as return

  int** Alocar(int **m,int rows,int cols) {  
  //^---tipo de retorno para int**
  
      m = (int **)malloc((rows)*sizeof(int*));
  
      for(int i=0; i<rows; i++) {
          m[i] = (int *)malloc(cols*sizeof(int));
      }
  
      return m; //retorna aqui
  }
  
  int main() {
      int **M = Alocar(M,100,100); //guarda o retornado em m
      Modificar(M,100,100);
  
      return 0;
  }
  

  Example in Ideone
  ^{I changed the amount to 10 to be easier to see}

• Passing the address of the matrix to the function:

  void Alocar(int ***m,int rows,int cols) {
      //----------^ agora int***
  
      *m = (int **)malloc((rows)*sizeof(int*)); //com * para ser o valor apontado
  
      for(int i=0; i<rows; i++) {
          (*m)[i] = (int *)malloc(cols*sizeof(int)); //com * tambem
      }
  }
  
  int main() {
      int **M;
  
      Alocar(&M,100,100); //passa o endereço agora
      Modificar(M,100,100);
  
      return 0;
  
  }
  

  Example also in Ideone

• Exclusively in C++ has even the idiomatic passage by reference that simplifies even more. The compiler internally ends up treating it like a pointer to the original:

  void Alocar(int **&m,int rows,int cols) {
      //------------^ referencia aqui o resto igual ao que tinha originalmente
      m = (int **)malloc((rows)*sizeof(int*));
  
      for(int i=0; i<rows; i++) {
          m[i] = (int *)malloc(cols*sizeof(int));
      }
  }
  
  int main() { //o main mantem-se igual
      int **M;
  
      Alocar(M,100,100);
      Modificar(M,100,100);
  
      return 0;
  }
  

  Final example in Ideone