尝试使用.substring将字符串转换为char

Question

我正在尝试使用以下方法将1d字符串数组转换为2d字符数组：

    '''''''''''''''
    variables
    '''''''''''''''
    const int width = 20;
    const int height = 20;  
    char arena[width][height];
    string arenaline[height];

    '''''''''''''''
    setup
    '''''''''''''''
    arenaline[1] = "####################";
    arenaline[2] = "#..................#";
    arenaline[3] = "#..................#";
    arenaline[4] = "###...###..###...###";
    arenaline[5] = "#.......#..#.......#";
    arenaline[6] = "###...###..###...###";
    arenaline[7] = "#..................#";
    arenaline[8] = "###...###..###...###";
    arenaline[9] = "#.......#..#.......#";
    arenaline[10] = "#########..#########";
    arenaline[11] = "###..............###";
    arenaline[12] = "###.#####..#####.###";
    arenaline[13] = "###.####....####.###";
    arenaline[14] = "###.####....####.###";
    arenaline[15] = "#.....###..###.....#";
    arenaline[16] = "#.....###..###.....#";
    arenaline[17] = "#..#..###..###..#..#";
    arenaline[18] = "#.....###..###.....#";
    arenaline[19] = "#.....###..###.....#";
    arenaline[20] = "####################";

    '''''''''''''''
    conversion
    '''''''''''''''
for (int i = 0; i < height; i++)
{
    for (int j = 0; j < width; j++)
    {

        arena[j][i] = arenaline[i].substr(j,1);

    }

}

我希望它从子字符串转换为char，以便我可以在数组中使用它。

我不能使用字符串而不是字符，因为它会破坏数组输出到控制台的功能。

Answer 1

substr返回一个字符串。对于“单字符”字符串，字符串不会隐式转换为char。

解决字符串单个字符的正确方法是string :: operator []或string :: at（）。