Question

我正在学习编程和c ++，这是我玩的第一门语言。我正在尝试创建一个简单的程序来加密和解密输入，以便可以使用字符串进行练习。

一切正常，但是不知何故，当我再次运行该程序时，我最终会弹出旧输入。

例如，如果我输入“ greg”，然后解密，然后再次输入greg，则会得到两个“ greg”（已加密）-因此无法正确重置，它将继续添加新单词而无需重置。

#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;
}

Answer 1

程序将secretword用于加密的输出和解密的输入，并且永远不会清除它。哎呀。

与其清除变量，不如减小变量的范围，以使它们无需清除且不会冲突。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，它会影响性能，但是您要做的手动记账越少越好。仅在无法达到性能目标的情况下使代码复杂化，并且仅将通过概要分析证明的代码复杂化为严重问题。

简单的Cypher程序无法正确重置

1 个答案: