I am trying to use column major order to a 2d matrix which I am using as 1d.

 int N=3,R=2;
 for (int i=0;i<N;i++){
    for (int j=0;j<R;j++){

      //printf("\nX[%d]=%d\t",i+j*N,res[i+j*N]); //column major
     // printf("\nX[%d]=%d\t",i*R+j,res[i*R+j]); //row major


    }
    printf("\n");
  }

but the column major doesn not work as it should.

Also,if I want to substract only column elements from a matrix (2d which is used as 1d),how should I approach?
Because ,that's why I am trying to present my matrix as column major order and then use its columns.

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#include <iostream>

void printrowmajor(int* arr, int width, int height)
{
    for (int i = 0; i < height; ++i)
    {
        for (int j = 0; j < width; ++j)
            std::cout<<arr[i * width + j]<<" ";

        std::cout<<"\n";
    }
}

void printcolumnmajor(int* arr, int width, int height)
{
    for (int i = 0; i < width; ++i)
    {
        for (int j = 0; j < height; ++j)
            std::cout<<arr[i + j * width]<<" ";

        std::cout<<"\n";
    }
}

int main()
{
    int arr[12] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};

    printrowmajor(&arr[0], 4, 3);

    std::cout<<"\n";

    printcolumnmajor(&arr[0], 4, 3);
    return 0;
}

This will print:

0 1 2 3 
4 5 6 7 
8 9 10 11 

0 4 8 
1 5 9 
2 6 10 
3 7 11 

This.. will also print the same thing:

int main()
{
    int arr[3][4] =
    {
        {0, 1, 2, 3},
        {4, 5, 6, 7},
        {8, 9, 10, 11}
    };

    printrowmajor(&arr[0][0], 4, 3);

    std::cout<<"\n";

    printcolumnmajor(&arr[0][0], 4, 3);
    return 0;
}

I am using the right indices.

Can you tell me why these 2 things don't work?

1) I have 2 matrices ,I multiply them and I can printf as row major or column major (everything is fine until now).
Now,if I store the result matrix (which I received from multiplication) :

     ...
     //store as column major
       for (int i=0;i<rows;i++){
          for (int j=0;j<cols;j++){
           res[i+j*rows]=res[i*cols+j];

       }
       }

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


         printf("\nB[%d]=%d\t",i+j*rows,res[i+j*rows]);
           //printf("\nB[%d]=%d\t",i*cols+j,res[i*cols+j]);

        }

      }

then neither printf shows the right results.

2) Can you tell me why this doesn't show the right results for column major but only for row major?

    int main(int argc, const char* argv[]) {
      int rows=3;
      int cols=2;

      int *A=(int*)malloc(rows*rows*sizeof(int));
      int *B=(int*)malloc(rows*cols*sizeof(int));
      int *result=(int*)malloc(rows*cols*sizeof(int));

      A[0]=1;
      A[1]=2;
      A[2]=3;
      A[3]=4;
      A[4]=5;
      A[5]=6;
      A[6]=7;
      A[7]=8;
      A[8]=9;

      B[0]=5;
      B[1]=6;
      B[2]=7;
      B[3]=8;
      B[4]=4;
      B[5]=5;


     for (int i=0;i<rows;i++){
       for (int j=0;j<cols;j++){
         result[i*cols+j]=0;
         for (int k=0;k<rows;k++){
        result[i*cols+j]+=A[i*rows+k]*B[k*cols+j];

       }

       }
    }


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

         printf("\nB[%d]=%d\t",i+j*rows,result[i+j*rows]);  //column major

         //printf("\nB[%d]=%d\t",i*cols+j,result[i*cols+j]);  //row major


        }
        printf("\n");
      }


      return 0; 

     }

Hold on for a moment, here. Is this supposed to be a C++ program, or a C program? While most C code is also valid C++, as a rule in C++ you want to avoid using certain C idioms and functions, with malloc(), free() and the standard C I/O functions (e.g., printf()) near the top of the list. If you are writing a C program, it would be advisable to post your questions in the C forum, to avoid confusion.

On a related note, why are you allocating the arrays dynamically, when you are using fixed array sizes?

Yes, I am using C code and compile it as C.Sorry.
Ok, I missed that with the matrices.
I corrected to a fixed size but still the same problem.

Since you are coding ALL in C, why not start fresh in the C forum ... I think I have some ideas that you may find useful.

Ok, I posted here , thanks!

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