如何在c ++中增加字母？

Posted 2021-04-17

我正在用c ++创建一个Caesar Cipher，我无法弄清楚如何增加一个字母。

我需要每次将字母增加1并返回字母表中的下一个字母。像下面这样添加1到'a'并返回'b'。

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

答案

这段代码应该让你开始。 letter是一个char，而不是一系列chars也不是字符串。

static_cast确保'a' + 1的结果被视为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

当你到达'z'（或'Z'！）并祝你好运时要小心！

另一答案

它按原样工作，但因为添加将表达式提升为int，所以你想再将它转换回char，以便你的IOStream将它呈现为一个字符而不是一个数字：

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

Output: b

还要添加环绕逻辑（这样当letter[0]是z时，你设置为a而不是递增），并考虑大小写。

另一答案

letter ++有用吗？总而言之，char是一种数字类型，所以它会增加ascii code。但我相信它必须定义为char letter而不是数组。但要注意在'Z'中添加一个。你会得到'['= P

#include <iostream>

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

这似乎运作良好;）

另一答案

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

另一答案

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'。

另一答案

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输入你的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:
" << text << endl;
    //return 0 (because the the main function is an int so it must return an integer value
    return 0;

    }

注意：这称为caesar密码加密，它的工作原理如下：

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

我希望能帮助你

另一答案

你可以使用'a'+（（字母 - 'a'+ n）％26）;在'z'之后假设您需要'a'，即'z'+ 1 ='a'

    #include <iostream>

    using namespace std;

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

       return 0;
    }

看到这个https://stackoverflow.com/a/6171969/8511215

另一答案

将字母[n]转换为byte *并将其引用值增加1。