使用字符陣列中的C++

Question

我有8個不同的陣列，並且每個陣列中有8個字符創建矩陣，

 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); 

如何將此轉換成8×8矩陣？我想把這些值逐列放置，就好像每個數組都轉換爲矩陣的一列，如array1到column1，array2到column2等等，就像array1的值將被放置成矩陣[0] [0]一樣， ，矩陣[1] [0]，矩陣[2] [0]等.. 我認爲，需要做的事情是這樣的：

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 
     } 

有你有它那麼你的方式，每個1×8作爲一列，而不是一行來像前一個:)

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

你自己是相當正確的：

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 
     } 

希望這有助於，如果它不只是讓我知道，我很樂意提供幫助。

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]; 
    } 
}