Thread: Given a matrix, write a function which... (transposition)

Hi everyone!
I should resolve this exercise, but I think I have some problems.
"Given a matrix written as a monodimensional array, write a function which change this matrix into his transposed".

Note: A Matrix can be written as a multidimensional array ( A[i][j] ) or as a monodimensional array ( A[i+n+j] ). In this case, it is required to use the monodimensional way.

Here there is what I written:

Code:

void trasp (float *A, int m, int n){

/*m is the number of arrows, n is the number of columns*/


        float *C;
        C= (float*)calloc(m*n, sizeof(float));


        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){

                C[j*m+i] =A[i*n+j];


            }
        }
        C=A;
        free(C);
}

I think that there are some mistakes, because it is not working.
The "multidimensional" way is easier.
This one should be easy too, but, I don't know why, I'm having some problems.

Thanks everyone!

Originally Posted by GReaper

What do you think these two lines do?

After I created the transposed of A (C), I assign C to A.
Doing this, the matrix given as input will be actually transformed into its transposed.
After doing that, I delete C, which becomes useless.

Originally Posted by GReaper

But you don't. Assignment goes from right to left, here you assign the pointer of matrix A to the pointer of matrix C, and then free/delete C(which is now A). If you wanted to change the 'A' matrix you need to take it as a double pointer, like this:

Firstable, thank you for your advices.

Unluckly, the code is still not working.
Here there is the function:

Code:

void traspo (float *A, int m, int n){


        float *C;
        C= (float*)calloc(m*n, sizeof(float));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                C[m*j+i] =A[n*i+j];


            }
        }
        free(A);
        A=C;
}

My matrix to transpose has two arrows and three columns.
In the first arrow it has -> 0 , 1 , 2
In the sec. arrow it has-> 0 , 2 , 4

When I try to transpose it, unluckly, the second arrow of the transposed matrix isn't correct.
I send you the code where there is written also the creation of the initial matrix, and the printf (the function transpo is the one I written above).

Code:

void traspo (float *A, int m, int n);


void main(){
    
    float *Q;
    int a=2;
    int c=3;
    
    Q= (float*)calloc (r*c, sizeof(float));


    int i, j;


        for (i=0; i<a; i++){
            for (j=0; j<c; j++){
                
                Q[i*c+j]=(float)(i+1)*j;
                printf("the value of the arrow %d and column %d of the matrix Q is %f\n\n", i+1,j+1,Q[i*c+j]);




            }


        }
        
        printf("\n\n\n I TRANSPOSE Q MATRIX \n\n\n");
        traspo(Q, a, c);
        for (i=0; i<c; i++){
                for (j=0; j<a; j++){
                    printf("the value of the arrow %d and column %d of the matrix Q is %f\n\n", i+1,j+1,Q[i*a+j]);
                }
        }




}

The transposed matrix should be like this one:

first arrow -> 0 , 0
sec. arrow -> 1 , 2
thir. arrow -> 2 , 4

But, instead, it is like this:


first arrow -> 0 , 0
sec.arrow -> 2 , 0
thir. arrow -> 2 , 4

Originally Posted by GReaper

But you don't. Assignment goes from right to left, here you assign the pointer of matrix A to the pointer of matrix C, and then free/delete C(which is now A). If you wanted to change the 'A' matrix you need to take it as a double pointer, like this:

ALL FINE GReaper, I solved that!

Thank you for your time!

By the way, the main function returns int, not void.