我想找到使用std::next_permutation解决转置密码的所有排列。为此，我创建了一个带有文本的2D动态字符数组。我有一个键指的是数组的列数。
离。
与key = 3
1 2 3
4 5 6
7 8 9
可能的排列将是
2 1 3
5 4 6
8 7 9
等
一切正常，直到<Unable to read memory> 0xfdfdfdfd <Error reading characters of string.>始终与transArr[4][0]的{{1}}相同。{/ p>
代码：
key = 6我无法理解为什么会发生这种情况，因为我在数组内部进行迭代，一切都已就绪。
我正在测试的文字是void transpositionCipher(char *text){
int key = 6, col = 0, row = 0, tempTextLength = length, numOfCol = 0, counter = 0,
i=0,j=0;
char **tranArr;
string fileName;
ofstream output,transpositions[5];
output.open("output.txt");
//for (key = 2; key < 7; key++){
output << "KEY IS " << key << "\n";
//fix length to create the columns.
while (tempTextLength % key != 0){
if (tempTextLength - 1 % key == 0){
tempTextLength--;
}
else tempTextLength++;
}
output << tempTextLength << "\n";
//fixed column length for each key.
numOfCol = tempTextLength / key;
//allocation
tranArr = new char*[key];
for (j = 0; j < key; ++j)
tranArr[j] = new char[numOfCol];
//fill the array with the text
for (i = 0; i < tempTextLength; i++){
if ((text[i] > 96 && text[i] < 123) || i < tempTextLength ){
if (col == numOfCol){
row++;
col = 0;
}
if (i >= length){
tranArr[row][col] = '-';
}
else tranArr[row][col] = text[i];
col++;
}
}
//create the permutations
fileName = "trans_key_"+to_string(key)+".txt";
transpositions[key-2].open(fileName);
transpositions[key-2] << "Permutations with key = " << key << "\n";
sort(tranArr, tranArr + key);
do {
for (int k = 0; k < numOfCol; k++){
for (int l = 0; l < key; l++){
transpositions[key-2] << tranArr[l][k] << '\t';
}
transpositions[key-2] << "\n";
}
transpositions[key-2] << "\n\n";
counter++;
} while (next_permutation(tranArr, tranArr + key));
//deallocation
for (j = 0; j < key; j++)
delete[] tranArr[j];
delete [] tranArr;
//reset
row = 0; col = 0; tempTextLength = length; counter = 0, numOfCol = 0;
//}
}