c++ - 通过计算差异来反转字符串中各个字符的字母值

Question

我正在尝试编写一个小函数，将小写字符翻转为字母表后半部分的对称对应字符 - 26 个字母 = 13/13。

a = z, b = y, c = x...

我试过下面的代码，但出于某种原因它只对第一个字符有效。

假设我输入“bamba”；它首先将“b”切换为“y”，但随后卡住并将所有其他字符也替换为“y”，我得到“yyyyy”。

我尝试了一些代码，发现如果我删除当前字符的依赖性，我可以安全地增加所有字母，比如说，1 (a = b, b = c...)

symmetric_difference = 1; **commented out** //21 - toCrypt[i];

我四处寻找，我找到的最接近的东西是“Rolling alphabet value of individual characters in string”但是它描述了一种看起来奇怪和多余的方式。

谁能告诉我我做错了什么(假设我做错了)？

#include <iostream>
using namespace std;

void crypto(char[]);

int  main()
{
    char toCrypt[80];

    cout << "enter a string:\n";
    cin >> toCrypt;

    crypto(toCrypt);

    cout << "after crypto:\n";
    cout << toCrypt;
}

void crypto(char toCrypt[]) // "Folding" encryption.
{
    int size = strlen(toCrypt);
    int symmetric_difference;

    for (int i = 0; i < size; i++)
    {
        symmetric_difference = 121 - toCrypt[i];    // Calculate the difference the letter has from it's symmetric counterpart.

        if (toCrypt[i] >= 97 && toCrypt[i] <= 110)  // If the letter is in the lower half on the alphabet,
            toCrypt[i] += symmetric_difference; // Increase it by the difference.
        else
        if (toCrypt[i] >= 111 && toCrypt[i] <= 122) // If it's in the upper half,
            toCrypt[i] -= symmetric_difference; // decrease it by the difference.
    }
}

Answer 1

你可以试试这个

for (int i = 0; i < size; i++)
{
   toCrypt[i] = 'z' - toCrypt[i] + 'a';
}