Transfering a 2-dimensional array into a single dimension array

Question

I'm trying to make a program in C that transfers a 2-dimensions-array(a matrix to be particular) into a single-dimension-array. For example, if we have a matrix with L lines and C columns, it should turn into aa single line newL=L*C. Therefore, if the matrix has 3 lines and 4 columns, the new array will have 3*4=12 as its size.

The matrix should turn to this:

1 2
     --> 1 2 3 4
3 4 

The problem I'm facing right now, is how to assign the matrix to the array without having random values or repeated values.

The piece of code I'm concerned with, goes like this:

for(k=1;k<=t;k++)
    {
        for(i=1;i<=l;i++)
        {
            for(j=1;j<=c;j++)
            {
                v[k]=m[i][j];
            }
        }
    }

k,i and j are counters of the matrix(2-dimensions-array) and the the array. two of which; i and j, are counters for the matrix and k is the array's counter. Notice that each one of them starts from 1 and goes to its size and in this size I will use 2 lines and 2 columns for the matrix therefore the array will have a size of 4(2*2).

• l is the number of lines in the array.
• c is the number of colunms in the array.
• t is the size of the array. t=l*c

Executing the code gives me this as a return:

1 2
    --> 4 4 4 4
3 4

Simply said, the piece of code will ALWAYS give the last value of the matrix to the array. So if I replace 4 with 5 in the matrix, the array will have 5 5 5 5.

EDIT:

Here is the full code to understand what I'm trying to do:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int c,i,j,l,k,t;
    printf("Donner le nombres des lignes: ");
    scanf("%d",&l);
    printf("Donner le nombres des colonnes: ");
    scanf("%d",&c);
    int m[l][c];
    t=l*c;
    int v[t];
    for(i=0;i<l;i++)
    {
        for(j=0;j<c;j++)
        {
            printf("Donner m[%d][%d]: ",i+1,j+1);
            scanf("%d",&m[i][j]);
        }
    }
    for(i=0;i<l;i++)
    {
        for(j=0;j<c;j++)
        {
            printf("%d\t",m[i][j]);
        }
        printf("\n");
    }
    printf("\n\n\n\n");
    for(k=1;k<=t;k++)
    {
        for(i=1;i<=l;i++)
        {
            for(j=1;j<=c;j++)
            {
                v[k]=m[i][j];
            }
        }
    }
    for(k=0;k<t;k++)
    {
        printf("%d\t",v[k]);
    }
    system("pause");
}

Thank you guys for the help, I found the correct way to do it.

Answer 1

1. You need not the outer loop
2. Array indices are zero-based in C

Thus, we have:

    for(k = 0, i = 0; i < o; i++)
    {
        for(j = 0; j < p; j++)
        {
            v[k++] = m[i][j];
        }
    }

where o and p - dimensions of the matrix m

Answer 2

If we have a multidimensional array like this:

int nums[3][3];

And we have:

int all[9];

And we've got:

int a, b;

We'll reference each of the nums like this:

nums[a][b];

Now think of what the values of a and b will actually be:

for (a = 0; a < 3; a++) {
   for (b = 0; b < 3; b++)
      all[((a * 3) + b)] = nums[a][b];
}

This will work so long as you multiply a with the number of elements it will iterate:

int nums[5][5];
int all[25];

int a, b;

for (a = 0; a < 5; a++) {
    for (b = 0; b < 5; b++)
        all[((a * 5) + b)] = nums[a][b];
}

Answer 3

You mention your question is "how to I fix the code?" I think plenty of people have given you the correct answer. This is your code along with the corrected code.

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int c,i,j,l,k,t;
    printf("Donner le nombres des lignes: ");
    scanf("%d",&l);
    printf("Donner le nombres des colonnes: ");
    scanf("%d",&c);
    int m[l][c];
    t=l*c;
    int v[t];
    for(i=0;i<l;i++)
    {
        for(j=0;j<c;j++)
        {
            printf("Donner m[%d][%d]: ",i+1,j+1);
            scanf("%d",&m[i][j]);
        }
    }
    for(i=0;i<l;i++)
    {
        for(j=0;j<c;j++)
        {
            printf("%d\t",m[i][j]);
        }
        printf("\n");
    }
    printf("\n\n\n\n");

    /* corrected code below */
    k = 0;
    for(i=0;i<l;i++)
    {
        for(j=0;j<c;j++)
        {
            v[k]=m[i][j];
            k++;
        }
    }
    /* corrected code above */

    for(k=0;k<t;k++)
    {
        printf("%d\t",v[k]);
    }
    system("pause");
}

Answer 4

This:

for(k=1;k<=t;k++)
   for(i=1;i<=l;i++)
      for(j=1;j<=c;j++)
          v[k]=m[i][j];

does not do what you think. Think about when you first loop through the j part, you will be setting all the 0th element of v the entire time, finally the last value you set will stick (ie, the one in position 1, 1 which happens to be 4 ). Then you will increment k to 1 and repeat it again, resulting in all 4's . You want this:

 for(i = 0; i < l; i++)
     for(j = 0; j < c; j++)
         v[i*l+j] = m[i][j]; // i*l + j gives you the equivelent position in a 1D vector. 

Make sure your v vector is the right size ie. int v[l*c]; . Also remember that in c zero indexing is used.If you really do need 1 based indexing (which you dont ...) then do this:

int k = 1;
  for(i = 1; i <= l; i++)
     for(j = 1; j <= c; j++)
         v[k++]=m[i][j];

But remember that this will make any further operations on this vector Gross . So dont do this ....

Answer 5

As long as the new array is the correct size, something like the following should work:

k=0;
for(i=0;i<l;i++){
    for(j=0;j<c;j++){
        v[k]=m[i][j];
        k++;
    }
}

Essentially, you are traversing over the matrix (your lines and columns--as you put it) and at the same time increasing the position (k) in the new array where you want that value to be put.

Answer 6

If you just want to access the matrix elements as a single dimension array, you could declare an int pointer v :

int m[3][4];
int *v = (int*)m;
// then access for example m[1][1] as v[5]

Or, to actually copy the array, use a double for (as in the other answers), a single for like below

int vv[12];
for(i = 0; i < 12; i++)
    vv[i] = m[i/4][i%4];

or just use memcpy :

memcpy(vv, m, 12*sizeof(int));