简单的Cypher程序无法正确重置

Simple Cypher Program Not Resetting Correctly

本文关键字：Cypher 程序 简单      更新时间：2023-10-16

我正在学习编程和C 作为我的第一语言。我正在尝试创建一个简单的程序来加密和解密输入，以便我可以使用字符串练习。

一切正常，但是以某种方式我第二次运行程序时会弹出旧输入。

例如，如果我输入" greg"，然后解密它，然后再次输入格雷格，我会得到两个" greg"(加密( - 因此，它无法正确重置，它将不重置新单词不重置。

#include <iostream>
#include <string>
using namespace std;
int main() {
  string alphabet{"a b c d e f g h i j k l m n o p q r s t u v w x y z 1 2 3 4 "
                  "5 6 7 8 9 0"};
  string key{"m h i p g f b t x u r l w a j e q k z d y v o c n s 9 5 0 7 8 2 "
             "3 4 1 6"};
  string word{};
  string secretword{};
  string decryptedword{};
  int selection{};
  // int newpos {};
  int position{};
  //int keypos{};
  do {
    cout << "-------------------------------------" << endl;
    cout << "Select an Option: " << endl;
    cout << "1: Encrypt" << endl;
    cout << "2: Decrypt" << endl;
    cout << "3: Quit" << endl;
    cin >> selection;
    if (selection == 1) {
      cout << "Enter a Word to Encrypt:";
      cin.sync();
      getline(cin, word);
      for (auto i : word) {
        if (isupper(i)) {
          cout << "Please use lower case only" << endl;
          break;
        }
        position = alphabet.find(i);
        secretword += key.at(position);
      }
      cout << "Encrypted Word: " << secretword << endl;
      secretword = "";
    }
    if (selection == 2) {
      cout << "Enter a Word to Decrypt: ";
      cin.sync();
      getline(cin, secretword);
      for (auto i : secretword) {
        if (isupper(i)) {
          cout << "Please use lower case only" << endl;
          break;
        }
        position = key.find(i);
        decryptedword += alphabet.at(position);
      }
      cout << "Decrypted Word: " << decryptedword << endl;
      decryptedword = "";
    }
  } while (selection != 3);
  return 0;
}

该程序使用secretword进行加密的输出和解密的输入，并且从未清除。哎呀。

而不是清除变量，而是减少其范围，以免它们被清除，也不能冲突。例如：

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string alphabet { "a b c d e f g h i j k l m n o p q r s t u v w x y z 1 2 3 4 "
                      "5 6 7 8 9 0" };
    string key { "m h i p g f b t x u r l w a j e q k z d y v o c n s 9 5 0 7 8 2 "
                 "3 4 1 6" };
    //int keypos{};
    int selection;
    do
    {
        cout << "-------------------------------------" << endl;
        cout << "Select an Option: " << endl;
        cout << "1: Encrypt" << endl;
        cout << "2: Decrypt" << endl;
        cout << "3: Quit" << endl;
        cin >> selection;
        if (selection == 1)
        {
            cout << "Enter a Word to Encrypt:";
            cin.sync();
            string word;
            string secretword;
            getline(cin, word);
            for (auto i : word)
            {
                if (isupper(i))
                {
                    cout << "Please use lower case only" << endl;
                    break;
                }
                auto position = alphabet.find(i);
                secretword += key.at(position);
            }
            cout << "Encrypted Word: " << secretword << endl;
            secretword = "";
        }
        if (selection == 2)
        {
            cout << "Enter a Word to Decrypt: ";
            cin.sync();
            string secretword;
            string decryptedword;
            getline(cin, secretword);
            for (auto i : secretword)
            {
                if (isupper(i))
                {
                    cout << "Please use lower case only" << endl;
                    break;
                }
                auto position = key.find(i);
                decryptedword += alphabet.at(position);
            }
            cout << "Decrypted Word: " << decryptedword << endl;
            decryptedword = "";
        }
    } while (selection != 3);
    return 0;
}

请注意，这并不总是正确的答案，有绩效的影响，但是您必须做的手动书籍越少越好，越好。仅当您无法遇到性能目标并使其复杂化的代码复杂化，而仅使分析证明是一个重大问题所证明的代码复杂。

使代码复杂化。
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