create matrix using character arrays in c++ - c++

Question

I have 8 different arrays, and each array has 8 characters in it,
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);
how to convert this into 8x8 matrix? I want to put these values column by column, like, each array is converted to one column of the matrix, like array1 to column1, array2 to column2, and so on, like array1 values will be placed like matrix[0][0], matrix[1][0], matrix[2][0] and so on..
I think, something like this needs to be done:
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
}
There you have it then your ways,each 1x8 coming as a COLUMN instead of a row as in previous one :)

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

You yourself were quite correct :
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
}
Hope this helps, if it doesn't just let me know, I'd be happy to help.

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];
}
}