在 C++ 中使用字符数组创建矩阵

Question

我有8个不同的数组，每个数组里面有8个字符，

        std::string str_v1 = v1.to_string();
        char arr_v1[9] = {0}; 
        std::copy(str_v1.begin(), str_v1.end(), arr_v1); // from str_v1 to str_v8

        std::string str_v8 = v8.to_string();
        char arr_v8[9] = {0}; 
        std::copy(str_v8.begin(), str_v8.end(), arr_v8); 

如何将其转换为 8x8 矩阵？我想逐列放置这些值，比如，每个数组都转换为矩阵的一列，比如array1到column1，array2到column2，依此类推，比如array1值将像matrix[0][0]一样放置, 矩阵[1][0], 矩阵[2][0] 等等.. 我认为，需要做这样的事情：

char matrix[8][8];
        for( int y=0;y<8;y++)
        {
                matrix[y][0] = arr_v1[y];
                matrix[y][1] = arr_v2[y];
                matrix[y][2] = arr_v3[y];
                matrix[y][3] = arr_v4[y];
                matrix[y][4] = arr_v5[y];
                matrix[y][5] = arr_v6[y];
                matrix[y][6] = arr_v7[y];
                matrix[y][7] = arr_v8[y];
        }

Answer 1

for( int y=0;y<8;y++)
        {
            for( int z=0;z<8;z++)
            {
             switch(y)
             {
             case 0:
             matrix[z][y] = arr_v1[z]; //Be pretty sure, possibly you are better than you believe
             break;                    // I've placed y before z as y is the outer loop, hence it 
             case 1:                   // should be responsible for ROWS and z for COLUMNS
             matrix[z][y] = arr_v2[z]; // Goes in matrix[0][1],[1][1],[2][1],[3][1]...[7][1]
             break;
             case 2:
             matrix[z][y] = arr_v3[z]; // Goes in matrix[0][2],[1][2],[2][2],[3][2]...[7][2]
             break;
             case 3:
             matrix[z][y] = arr_v4[z]; // Goes on
             break;
             case 4:
             matrix[z][y] = arr_v5[z]; // And on
             break;
             case 5:
             matrix[z][y] = arr_v6[z]; // And on
             break;
             case 6:
             matrix[z][y] = arr_v7[z];
             break;
             case 7:
             matrix[z][y] = arr_v8[z];
             break;
             }
            }  // Finally all 8 1x8 arrays stored into single 8x8 matrix
        }

你有它然后你的方式，每个 1x8 以 COLUMN 的形式出现，而不是像前一个那样的一行 :)

Answer 2

char *matrix[9];
for (int i = 0; i < 9; ++i) {
    matrix[i] = new char[9];
    std::copy(your_ith_string.begin(), your_ith_string.end(), matrix[i]);
}
//Finish your work with the matrix
for (int i = 0; i < 9; ++i) {
    delete[] matrix[i];
}

Answer 3

你自己说的很对：

for( int y=0;y<8;y++)
        {
            for( int z=0;z<8;z++)
            {
             switch(y)
             {
             case 0:
             matrix[y][z] = arr_v1[z]; //Be pretty sure, possibly you are better than you believe
             break;                    // I've placed y before z as y is the outer loop, hence it 
             case 1:                   // should be responsible for ROWS and z for COLUMNS
             matrix[y][z] = arr_v2[z]; // Goes in matrix[1][0],[1][1],[1][2],[1][3]...[1][7]
             break;
             case 2:
             matrix[y][z] = arr_v3[z]; // Goes in matrix[2][0],[2][1],[2][2],[2][3]...[2][7]
             break;
             case 3:
             matrix[y][z] = arr_v4[z]; // Goes on
             break;
             case 4:
             matrix[y][z] = arr_v5[z]; // And on
             break;
             case 5:
             matrix[y][z] = arr_v6[z]; // And on
             break;
             case 6:
             matrix[y][z] = arr_v7[z];
             break;
             case 7:
             matrix[y][z] = arr_v8[z];
             break;
             }
            }  // Finally all 8 1x8 arrays stored into single 8x8 matrix
        }

希望这会有所帮助，如果它不只是让我知道，我很乐意提供帮助。

Answer 4

char matrix[8][8];
char *arr[8] = {arr_v1, arr_v2, arr_v3, arr_v4, arr_v5, arr_v6, arr_v7, arr_v8};
for( int y=0;y<8;y++) {
    for (int i = 0; i < 8; ++i) {
        matrix[y][i] = arr[i][y];
    }
}