我如何在 C++ 中增加字母？

How do i increment letters in c++?

提问by Marobri

I'm creating a Caesar Cipher in c++ and i can't figure out how to increment a letter.

我正在用 C++ 创建一个凯撒密码，我不知道如何增加一个字母。

I need to increment the letter by 1 each time and return the next letter in the alphabet. Something like the following to add 1 to 'a'and return 'b'.

我需要每次将字母加 1 并返回字母表中的下一个字母。类似于以下的内容将 1 添加到'a'并返回'b'.

char letter[] = "a";
cout << letter[0] +1;

采纳答案by Johnsyweb

This snippet should get you started. letteris a charand not an array of chars nor a string.

这个片段应该让你开始。letter是一个char而不是一个chars数组，也不是一个字符串。

The static_castensures the result of 'a' + 1is treated as a char.

该static_cast确保的结果'a' + 1被视为一个char。

> cat caesar.cpp          
#include <iostream>

int main()
{
    char letter = 'a';
    std::cout << static_cast<char>(letter + 1) << std::endl;
}

> g++ caesar.cpp -o caesar
> ./caesar                
b

Watch out when you get to 'z'(or 'Z'!) and good luck!

当你到达'z'（或'Z'！）时要小心，祝你好运！

回答by Lightness Races in Orbit

It works as-is, but because the addition promotes the expression to intyou want to cast it back to charagain so that your IOStream renders it as a character rather than a number:

它按原样工作，但因为添加将表达式提升为int您希望char再次将其强制转换，以便您的 IOStream 将其呈现为字符而不是数字：

int main() {
   char letter[] = "a";
   cout << static_cast<char>(letter[0] + 1);
}

Output: b

输出：b

Also add wrap-around logic (so that when letter[0]is z, you set to arather than incrementing), and consider case.

还添加环绕逻辑（以便 when letter[0]is z，您设置为a而不是递增），并考虑大小写。

回答by thim

Does letter++ work? All in all char is a numeric type, so it will increment the ascii code. But I believe it must be defined as char letternot an array. But beware of adding one to 'Z'. You will get '[' =P

字母++有用吗？总而言之 char 是数字类型，因此它会增加ascii 代码。但我相信它必须被定义为char letter不是一个数组。但要注意在“Z”上加一个。你会得到 '[' =P

#include <iostream>

int main () {
    char a = 'a';
    a++;
    std::cout << a;
}

This seems to work well ;)

这似乎运作良好;)

回答by Roee Gavirel

char letter = 'a'; 
cout << ++letter;

回答by Kurt

It works but don't forget that if you increment 'z' you need to get 'a' so maybe you should pass by a check function that output 'a' when you get 'z'.

它有效，但不要忘记，如果你增加“z”，你需要得到“a”，所以也许你应该通过一个检查函数，当你得到“z”时输出“a”。

回答by Waleed Qutob

waleed@waleed-P17SM-A:~$ nano Good_morning_encryption.cpp waleed@waleed-P17SM-A:~$ g++ Good_morning_encryption.cpp -o Good_morning_encryption.out waleed@waleed-P17SM-A:~$ ./Good_morning_encryption.out Enter your text:waleed Encrypted text: jnyrrq waleed@waleed-P17SM-A:~$ cat Good_morning_encryption.cpp

waleed@waleed-P17SM-A:~$ nano Good_morning_encryption.cpp waleed@waleed-P17SM-A:~$ g++ Good_morning_encryption.cpp -o Good_morning_encryption.out waleed@waleed-P17SM-A:~$ ./Good_morning_encryption.out text:waleed 加密文本：jnyrrq waleed@waleed-P17SM-A:~$ cat Good_morning_encryption.cpp

    #include <iostream>
    #include <string>

    using namespace std;


    int main() {

    //the string that holds the user input
    string text;
    //x for the first counter than makes it keeps looping until it encrypts the user input
    //len holds the value (int) of the length of the user input ( including spaces)
    int x, len;

    //simple console output
    cout << "Enter your text:";
    //gets the user input ( including spaces and saves it to the variable text
    getline(cin, text);
    //give the variable len the value of the user input length
    len = (int)text.length();
    //counter that makes it keep looping until it "encrypts" all of the user input (that's why it keeps looping while its less than len
    for(x = 0; x < len; x++) {
    //checks each letts (and spaces) in the user input (x is the number of the offset keep in mind that it starts from 0 and for example text[x] if the user input was waleed would be w since its text[0]
    if (isalpha(text[x])) {
    //converts each letter to small letter ( even though it can be done another way by making the check like this if (text[x] =='z' || text[x] == 'Z')
    text[x] = tolower(text[x]);
    //another counter that loops 13 times
    for (int counter = 0; counter < 13; counter++) {

    //it checks if the letts text[x] is z and if it is it will make it a
    if (text[x] == 'z') {

    text[x] = 'a';

    } 
    //if its not z it will keeps increamenting (using the loop 13 times)
    else {


    text[x]++;

    }

    }
    }
    }
//prints out the final value of text
    cout << "Encrypted text:\n" << text << endl;
    //return 0 (because the the main function is an int so it must return an integer value
    return 0;

    }

Note: this is called caeser cipher encryption it works like this :

注意：这称为 caeser 密码加密，其工作方式如下：

ABCDEFGHIJKLMNOPQRSTUVWXYZ NOPQRSTUVWXYZABCDEFGHIJKLM so for example my name is waleed it will be written as : JNYRRQ so its simply add 13 letters to each letter

ABCDEFGHIJKLMNOPQRSTUVWXYZ NOPQRSTUVWXYZABCDEFGHIJKLM 所以例如我的名字是 waleed 它将被写成：JNYRRQ 所以它只需在每个字母上添加 13 个字母

i hope that helped you

我希望对你有帮助

回答by Bimalendu

You can use 'a'+((letter - 'a'+n)%26);assuming after 'z' you need 'a' i.e. 'z'+1='a'

您可以使用'a'+((letter - 'a'+n)%26); 假设在 'z' 之后你需要 'a' 即 'z'+1='a'

    #include <iostream>

    using namespace std;

    int main()
    {
       char letter='z';
       cout<<(char)('a' + ((letter - 'a' + 1) % 26));

       return 0;
    }

See this https://stackoverflow.com/a/6171969/8511215

请参阅此https://stackoverflow.com/a/6171969/8511215

回答by belgther

cast letter[n] to byte* and increase its referenced value by 1.

将 letter[n] 转换为 byte* 并将其引用值增加 1。