Thread: quick question about creating a matrix using the new operator

  1. #1
    Registered User
    Join Date
    Dec 2004

    Unhappy quick question about creating a matrix using the new operator


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

          1 #include<iostream>
          2 using namespace std;
          3 int main()
          4 {
          7         int** M;
          8          M= new int* [5];
          9         for(int i=0; i<5; i++)
         11                 for(int j=0; j<5; j++)
         12                 M[i][j]=5;
         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 }
    Not to know, not to learn is shame

  2. #2
    Registered User Micko's Avatar
    Join Date
    Nov 2003
    Because you write in memory you didn't allocate.
    Probably this is what you want. Notice the difference?
    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++)
         for(int k=0; k<5; k++,cout<<endl)
                 for(int m=0; m<5; m++)
                         cout<<M[k][m]<<" ";
         return 0;
    - Micko
    Gotta love the "please fix this for me, but I'm not going to tell you which functions we're allowed to use" posts.
    It's like teaching people to walk by first breaking their legs - muppet teachers! - Salem

  3. #3
    Registered User
    Join Date
    Apr 2003
    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]
    Last edited by 7stud; 04-27-2005 at 11:50 PM.

  4. #4
    Chief Code Coloniser!
    Join Date
    Apr 2005
    And don't forget to free your memory up afterwards using delete [].

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