How can I add a given amount of 2D vectors into a 3D vector?

Question

I am currently developing a maze generator and splitting it up into cells which I aim to add up to create a maze, and each cell is a 2d vector where classes are employed. How can I add all of the corresponding 2d vectors to a 3d vector to generate the maze? Below is the code I have been employing.

std::vector<std::vector<std::vector<char> > > maze::matrix (int rows, int columns, std::vector<std::vector<char> > cell)  {

    std::vector<std::vector<std::vector<char> > > maze;

    for (int i = 0; i < rows; i++) {
        
        maze.push_back(std::vector<std::vector<char> >());

        for (int j = 0; j < columns; j++) {

            maze.at(i).push_back(cell);

        }
    }

    return maze;

}

Answer 1

First, as suggested, you should make aliases for the types, so that you can more clearly see the issue:

#include <vector>

// Create aliases
using Char1D = std::vector<char>;
using Char2D = std::vector<Char1D>;
using Char3D = std::vector<Char2D>;

int main()
{
  // Sample set of cells 
  Char2D cells = {{'x','y'},{'0','1'}};
  Char2D cells2 = {{'0','1'},{'x','y'}};

  // The maze to add the above cells
  Char3D maze;

 // Now add the cells to the maze
  maze.push_back(cells);
  maze.push_back(cells2);
}

That code adds 2 different Char2D cells to the maze .

The issue with your code is that you were basically calling push_back with the wrong types -- you were calling maze[i].push_back , but maze[i] is already a Char2D , so you were trying to push_back a Char2D into a Char2D .

More than likely, your code was not following your specification of adding 2D vectors to the 3D vector.

Answer 2

Please also include the unexpected behavior you get. In this case, this code won't compile.

maze is a vector<vector<vector<char>>> ie 3D vector

maze.at(i) is a vector<vector<char>> ie 2D vector

You are trying to push_back(cell), where cell is a 2D vector.std::vector::push_back takes an element of the vector, so for a 2D vector it takes a 1D vector ie maze.at(i).push_back(std::vector<char>{}) .