家庭作业（修订）：凯撒密码，当char是>'z'（c ++）时，环绕 [英] Homework(revision): Caesar cipher, wrap around when char's are >'z' (c++)

问题描述

我刚才在这里，手中添加了charci ascii引用来增加每个字母的密码转换。然而，我不知道如何解决字符高于'z'的问题。

有人可以给我一个提示，当角色到达时字母表的结尾。当然，我不希望有人为我做我的工作。

  char decrypt（char letter）
 {
 int increment = 9; 
 if（letter ==''）
 {
 return letter; 
} 
 letter + = increment; 
回信; 
} 
 
 int main（）
 {
 char message [446]; int i = 0; char space =''; 
 ifstream in（encryptedText.txt）; 
 if（in.getline（message，446））
 {
 while（message [i]）
 {
 cout<< decrypt（tolower（message [i]））<< ENDL; 
 i ++; 
} 
} 
 else 
 {cout<<< 无法读取文件<< endl;} 
 cout<< ENDL; 
 system（pause）; 
} 
  

解决方案

模运算是你的朋友。每当整数形成环而不是序列时，可以应用模运算，如：

  5％4 == 17％ 4 
  

您需要在相对于'a' ，当然可以适当地减去'a'。

I was here earlier and got a hand adding to chars ascii references to increment the cipher shift on each letter. However i have no idea how to fix the problem of characters being higher than 'z'.

Can someone give me a hint towards how to wrap around when the characters reach the end of the alphabet. I don't expect anyone to do my work for me, of course.

char decrypt(char letter)
{
int increment = 9;
if(letter == ' ')
{
    return letter;
}
letter += increment;
return letter;
}

int main()
{
char message[446]; int i = 0; char space = ' ';
ifstream in("encryptedText.txt");
if(in.getline(message, 446))
{
    while(message[i])
    {
            cout << decrypt(tolower(message[i])) << endl;
            i++;
    }
}
else
{cout << "Can't read file" << endl;}
cout << endl;  
system("pause");
}

解决方案

The modulo operation is your friend. Whenever integers form a ring instead of a sequence, a modulo operation can be applied, like:

5 % 4 == 17 % 4

You need to compute this in the integer space relative to 'a', of course, subtracting 'a' appropriately.

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