Question

I am currently browsing over some old college snippets of c++ code. Back then, one of the other class was assigned with doing a matrix class using double pointers and 2D arrays. Luckily (or unluckily upon hindsight) I never did get a chance to learn stuff like that. I borrowed their code when we graduated for future review. If anyone can please explain to me what exactly happens in this snippet:

  //This is a constructor of a 1x1 matrix
  signal::signal(){
     _nrows = 1;
     _ncols = 1;
     _coef = new double*[_nrows];
     _coef[0] = new double[_ncols];
     _coef[0][0] = 0.0;
  }

Just a sidenote, _coef is a ** of type double.

From what I understand, _nrows and _ncols are given a value of 1 (meaning their sizes). Then, the code dynamically creates a double* out in the heap with elements equal to _nrows; the problem is, I dont exactly know what happens next. Why is the array corresponding to _ncols not a pointer? Why is it assigned _coef[0]?

Answer 1

In memory, a two dimensional array (n, m) looks more or less like this

_coef -> | _coef[0] -> {1, 2, 3, ..., m}
         | _coef[1] -> {1, 2, 3, ..., m}
         | _coef[2] -> {1, 2, 3, ..., m}
         | ...
         | _coef[n] -> {1, 2, 3, ..., m}

_coef points to an array of n pointers. And each of these pointers point to an array of m doubles.

So, in your case _coef points to an array of 1 pointer and this pointer points to an array of one double.

Now to your questions

1. It is not a pointer, because in your second dimension, you finally want to store the doubles, not pointers.
2. It is assigned to _coef[0], because it is the first, and only, row of your two dimensional array.