Switching Matrix columns and rows in C++ - c++

Question

I have a 6 x 6 matrix, and I am storing its values in a one dimensional array of size 36. I want to rearrange it so that the rows are the columns and the columns are the rows. My method is trying to copy the values into another array, but sorted properly. I am trying a for loop:
for (int i = 0; i < 6; ++i){
copyArray[i]= array[i*6];
}
This works fine for the first new row made up of the first column, but how do I continue to do all of them? I have tried nested for loops but cannot come up with the proper algorithm using iterators. I could do this manually, but would like to have this done by code.
I am coding in C++, but if anyone can do it in a similar language that would be fine. I feel this is a math problem.
The Question: How do I solve for switching out the rows and columns?
(Example: if I am denoting the first rows and columns as 0, in a 6x6 matrix, then both rows and columns go from 0 to 5. Therefore, by switching rows and columns, the value in row 2, column 5 would be switched with the value in row 5, column 2.)

Answer 1

Couldn't you just use a nested for loop and do something like this?
for (int i = 0; i < 6; ++i)
for (int j = 0; j < 6; ++j)
copyArray[i*6+j]= array[j*6+i];
Here's a test program you can run to show it works:
#include <stdio.h>
int main()
{
int array[36] = {1,1,1,1,1,1,
2,2,2,2,2,2,
3,3,3,3,3,3,
4,4,4,4,4,4,
5,5,5,5,5,5,
6,6,6,6,6,6};
int copyarray[36];
for (int i = 0; i < 6; ++i)
for (int j = 0; j < 6; ++j)
copyarray[i*6+j]= array[j*6+i];
for (int i = 0; i < 36; ++i)
{
if (i % 6 == 0)
printf("\n");
printf("%d ", array[i]);
}
printf("\n");
printf("\n");
for (int i = 0; i < 36; ++i)
{
if (i % 6 == 0)
printf("\n");
printf("%d ", copyarray[i]);
}
return 0;
}
Output:
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6

Answer 2

You should keep your one-D array inside a class.
Then you can swap the dimensionality of the data without actually moving it.
class MySwapableArray
{
bool normal;
int array[36];
public:
int& operator()(int x, int y)
{
int h = normal ? x : y;
int v = normal ? y : x;
return array[h*6+v];
}
void swapRowsAndColumns()
{
normal = ! normal;
}
};
If you want to use operator[] like A[x][y] then see: How to overload array index operator for wrapper class of 2D array?

Answer 3

A template for doing so.
template<class X> std::unique_ptr<typename std::pointer_traits<X>::element_type[]>
transpose(const X& p, int x, int y) {
using T = std::pointer_traits<X>::element_type;
std::unique_ptr<T[]> r = new T[x*y];
for(int a = 0; a < x; ++a)
for(int b = 0; b < y; ++b)
r[x*a+b] = p[y*b+a];
return r;
}

Answer 4

If you want to avoid a double-for loop, you should be able to do it mathematically with something like
for (int i = 0; i < 36; ++i) {
copyarray[6 * (i % 6) + i / 6] = array[i]
}
Basically, 6 * (i % 6) aligns you to the right row in the transposed matrix, and the i / 6 (ab)uses integer division to translate the column value into a row value. floor (i / 6.0) is more correct if you can live with the clutter.

Answer 5

for( int i=0;i<6;i++)
for( int j=0;j<6;j++)
{ }//You can do whatever you want on original matrix here
Let's say that you want "transpose of matrix" and do some processing .Basically switch i and j
for( int j=0;j<6;j++) // replaceed i with j
for( int i=0;i<6;i++)
{ }//You can do whatever you want on transpose matrix here