quick question about creating a matrix using the new operator

Hi@all

Please can anyone tell me why when I run this program, I am getting memory fault error..

Code:

---

      1 #include<iostream>
      2 using namespace std;
      3 int main()
      4 {
      5
      6
      7        int** M;
      8          M= new int* [5];
      9        for(int i=0; i<5; i++)
    10
    11                for(int j=0; j<5; j++)
    12                M[i][j]=5;
    13
    14
    15        for(int k=0; k<5; k++)
    16                for(int m=0; m<5; m++)
    17                        cout<<M[k][m];
    18          return 0;
    19 }

---

Because you write in memory you didn't allocate.
Probably this is what you want. Notice the difference?

Code:

---

#include<iostream>
using namespace std;

int main()   
{
      int i,j;
      int** M;
      M= new int* [5];
      for(i=0; i<5; i++)
              M[i]= new int [5];
             
      for(i=0; i<5; i++)   
              for(j=0; j<5; j++)
                        M[i][j]=5;
   
   
    for(int k=0; k<5; k++,cout<<endl)
            for(int m=0; m<5; m++)
                    cout<<M[k][m]<<" ";
   
    return 0;
}

---

- Micko

This line:

M= new int* [5]

creates an array filled with int pointers:

[int*, int*, int*, int*, int*]

There are five elements, and you can access each element like this:

M[0], M[1], M[2], etc.

Now, what is M[0][2]? M is a single dimensional array, so there is no second index.

What you need to do is dynamically create 5 arrays of type int and assign each one to a pointer in M[]. After you do that, the notation M[0][1] will first access the pointer at index position 0 in M[]:

(M[0]) [1]

and since M[0] points to an array of ints, the second index specified will access the int at position 1:

(M[0]) [1]

And don't forget to free your memory up afterwards using delete [].