C++ 移动一串字符？
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Shifting a string of characters?
提问by Christian
I'm writing a program that solves Caesar ciphers in C++. It takes a string of the alphabet and shifts it to the left each loop: "abc....yz" --> "bcd.....yza". The problem is after another loop it goes: "bcd.....yza" --> "cde.....yzaa".
我正在编写一个用 C++ 解决凯撒密码的程序。它需要一个字母字符串并将其向左移动每个循环："abc....yz" --> "bcd.....yza"。问题是在另一个循环之后：“bcd.....yza”-->“cde.....yzaa”。
char temp; // holds the first character of string
string letters = "abcdefghijklmnopqrstuvwxyz";
while (true)
{
temp = letters[0];
for (int i = 0; i < 26; i++)
{
if (i == 25)
{
letters += temp;
}
letters[i] = letters[i + 1];
cout << letters[i];
}
cin.get();
}
Copy and paste that code and you'll see what I'm talking about. How do I fix this mysterious problem?
复制并粘贴该代码，您就会明白我在说什么。我该如何解决这个神秘的问题？
回答by MSalters
If I'm not mistaken, your loop does precisely the same as the following code:
如果我没记错的话，您的循环与以下代码完全相同：
letters = letters.substr(1,25) + letters.substr(0,1);
// [skip 1, take 25] + [first char goes last]
回答by Nemo
I think you need letters to be 27 characters, not 26, and instead of letters += temp(which grows the string every time), use letters[26] = temp[0].
我认为您需要字母为 27 个字符，而不是 26 个字符，而不是letters += temp（每次都会增加字符串），使用letters[26] = temp[0].
...at which point you can just ditch tempentirely:
...此时你可以temp完全放弃：
string letters = "abcdefghijklmnopqrstuvwxyz.";
while (true)
{
letters[26] = letters[0];
for (int i = 0; i < 26; i++)
{
letters[i] = letters[i + 1];
cout << letters[i];
}
cin.get();
}
[edit]
[编辑]
Although the more natural way to handle this is to use arithmetic on the characters themselves. The expression 'a' + ((c - 'a' + n) % 26)will shift a char cby nplaces Caesar-style.
尽管处理这个问题的更自然的方法是对字符本身使用算术。表达式'a' + ((c - 'a' + n) % 26)将转向一个烧焦c的n地方撒风格。
回答by yasouser
You can achieve this easily using valarray< char >::cshift(n)(cyclical shift) method.
您可以使用valarray< char >::cshift(n)（循环移位）方法轻松实现这一点。
