Saving a row of a vector into another vector

[imperialcplusplus]

Hello, I am learning c++ and I am currently at halt.

I am trying to write a function such that:
1) It takes in input a one dimensional vector and an integer which specifies a row.
2) The numbers on that row are put into an output vector for later use.

The only issue is that this online course states that I must use another function that I have made before that allows a 1d vector with one index be able to have two indexes.

it is:

int 2d_to_1d(int row, int col, int rowlen){
    return row*rowlen+col;
}

logically what I am trying to do:
1) I use this function to store the input vector into a temporary vector as a 2D matric with i as the x axis and y as the y axis.
2) from there I have a loop which reads out the numbers on the row needed and stores it in the output vector.

so far I have:

  void row(int r, const std::vector<int>& in, std::vector<int>& out){

    int rowlength = std::sqrt(in.size());
    std::vector <int> temp;

    for(int i = 0; i < rowlength; i++){           // i is vertical and j is horizontal
        for(int j = 0; j < side; j++){
           *** temp[twod_to_oned(i,j,side)];        

// now stored as a 2D array(matrix?)
        }
    }

    for(int i=r; i=r; i++){
        for(int j=0; j< rowlength; j++){
           *** out[temp[]]
        }

      }

I'm pretty sure there is something wrong in the first and last loop which turns into a 2D matric then stores the row.

I starred the parts that are incomplete due to my lack of knowledge.

How could I overcome this issue? I would appreciate any help, Many thanks.

[ne555]

> I starred the parts that are incomplete due to my lack of knowledge.
don't do that, I shouldn't have to modify your code in order to test it.

> temp[twod_to_oned(i,j,side)];
that does nothing
you simply try to access an element (which doesn't exist, by the way) but don't do anything with it, you don't check its value or modify it.


your yo-yo code is quite inefficient, the task should be O(n) but you make it O(n^2)
let's try to simplify it, set all the elements in a row to the value 42.