for (int i = 0; i<mdim_; i++)
{
for (int j = 0; j<ndim_; j++)
{
data_[j * ndim_ + i];
}
}

to change the matrix 1 3 2 4 stored in data, into row order. If I print data_[j * ndim_ + i]; to the screen then it will give me 1 2 3 4 however I can't figure out how to save this into a another vector, as this would involve introducing another loop which messes things up.

Can anyone please tell me if there is a way to do this?

Completely worthless description without context. Start at the beginning, not the tail end of the problem. Remember, we aren't there watching how you got to this point.

Since your 2D data is being stored in 1D arrays, all you need to do here is play indexing games. Suppose you had a matrix of MxN stored in a 1D array M*N in length. If the matrix is stored in column major order, and you need to convert it to row major order, you will have to do some copying. So, try this approach:

0. Given an MxN ( M=width, N=height ) Matrix ( mat0 ) stored in column major order in a 1D array ( arr0 ) of length M*N
1. Create another 1D ( arr1) array of the same size (M*N) that will contain the matrix when it is converted to row major order
2. Iterate over arr0 using an index value ( idx0 ) beginning at index 0 to M*N ( the length of the array )
3. Compute an index ( idx1 ) into the row major matrix the column major array index ( idx0 )
4. Store arr0[idx0] into arr1[idx1]

To do step 3, you should take advantage of integer division and modulo properties. Namely

Well that basically is the beginning...I've been trying for a few days now to append below one matrix to another. All initial matrices are stored in column order which makes this difficult.

Various methods I've tried haven't worked, so I came up with a new idea of transforming both matrices into row order, sticking them together, and transforming back into column order according to the number of rows of both matrices.

Hello there guys! I must implement a linked list in assembly (using MIPS). We're supposed to dynamically allocate memory for the nodes. This is what I am asked to do: ...