Question

我正在尝试自学C ++，这是我正在进行的一个程序，但我遇到了密码循环问题。甚至为什么我输入正确的用户名和密码，它只是一次又一次地问我用户名而不是去虚拟菜单功能

#include <iostream>
#include <stdio.h>
#include <string>
#include <fstream>

using namespace std;
void menu(int argc, const char * argv[], char text[2000]);

string encryptDecrypt(string toEncrypt) {
    char key = 'K'; //Any char will work
    string output = toEncrypt;

    for (int i = 0; i < toEncrypt.size(); i++)
        output[i] = toEncrypt[i] ^ key;

    return output;
}

void menu(int argc, const char * argv[], char text[2000])
{

    system("color 0A");
    //ifstream my_input_file;
    ofstream my_output_file;

    my_output_file.open("output_data.txt");

    cout<<"Please enter your text: ";
    //cin>>text;
    //cin.ignore();
    //cin.getline(text, sizeof text);
    cin.getline(text,2000);
    //cout<<"You entered: "<< text <<"\n";

    //string encrypted = encryptDecrypt("kylewbanks.com");
    string encrypted = encryptDecrypt(text);
    cout << "Encrypted:" << encrypted << "\n";

    string decrypted = encryptDecrypt(encrypted);
    cout << "Decrypted:" << decrypted << "\n";

    my_output_file << "Your encrypted text is: "  << encrypted;
    my_output_file.close();
    cin.get();
}

int main()
{
    string username;
    string password;
    do {
        cout << "username: ";
        getline(std::cin, username);
        if (username == "John") {
            std::cout << "password: ";
            getline(std::cin, password);
            if (password != "1234") {
                cout << "invalid password. try again." << std::endl;
            }
            else if (password == "1234"){
                void menu(int argc, const char * argv[], char text[2000]);
            }
        }
        else {
            std::cout << "invalid username. try again." << std::endl;
        }
    } while (password != "1234");
    return 1;
}

Answer 1

如果密码等于“cherry”，那么你什么也不做，只需在本地声明函数即可菜单

else if (password == "cherry"){
     void menu(int argc, const char * argv[], char text[2000]);
}

本声明

     void menu(int argc, const char * argv[], char text[2000]);

不是函数调用。这是一个功能声明。

但是如果输入“1234”，则循环结束，因为它的条件是

} while (password != "1234");

编辑：我看到你更新了帖子和替换语句

else if (password == "cherry"){

的

else if (password == "1234"){

然而，实质上没有任何改变。在新陈述之后，仍然有一个函数声明。

Answer 2

#include <iostream>
#include <stdio.h>
#include <string>
#include <fstream>

using namespace std; //the point of using namespace std is to avoid writing std:: every time
string encryptDecrypt(string toEncrypt) {
    char key = 'K'; //Any char will work
    string output = toEncrypt;

    for (int i = 0; i < toEncrypt.size(); i++)
        output[i] = toEncrypt[i] ^ key;

    return output;
}
// better use the string class rather than an array of chars. Made it a local variable.
void menu(int argc, const char * argv[])
{

    system("color 0A");
    //ifstream my_input_file;
    ofstream my_output_file;

    my_output_file.open("output_data.txt");

    cout << "Please enter your text: ";
    string text;
    cin >> text;

    string encrypted = encryptDecrypt(text);
    cout << "Encrypted:" << encrypted << "\n";

    string decrypted = encryptDecrypt(encrypted);
    cout << "Decrypted:" << decrypted << "\n";

    my_output_file << "Your encrypted text is: "  << encrypted;
    my_output_file.close();
}

int main(int argc, const char * argv[])
{
  string username;
  string password;
  //changed the do while to be
  while(password != 1234) {
     cout << "username: ";
     cin >> username;
     if (username == "John") {
        cout << "password: ";
        cin >> password;

        //swapped comparisons to avoid 1 comparison and make the code clearer
        if (password == "cherry")
          menu(argc, argv); //it didn't make sense giving text as an argument 
                            //because you then do a cin >> text in menu.

        else cout << "invalid password. try again." << endl;   

    }
    else cout << "invalid username. try again." << endl;
}

Answer 3

首先，不要使用using namespace std;。只是不要。这是不好的做法，不管其他人是否不指出你。其次，认真阅读函数的调用，制作和反馈参数，特别是如果你想复制别人的代码。即使你纠正了你的循环，如果你不知道哪些参数传递给你的menu函数，那么它就是zilch。

现在已经开始了，请参阅以下版本的代码。

#include <iostream>
#include <string>

void menu()
{
    std::cout << "I am inside the menu!" << std::endl;
}

int main()
{
    std::string username;
    std::string password;
    std::string access = "cherry";

    do {
        std::cout << "Username: ";
        std::getline(std::cin, username);
        if (username != "John") {
            std::cout << "Invalid username. Please try again.\n";
            continue;
        }
        std::cout << "Password: ";
        std::getline(std::cin, password);
        if (password != access){
            std::cout << "Incorrect password. Resetting access procedure.\n";
            continue;
        }
    } while (password != access);

    menu();
    std::cin.get();
    return 1;
}

了解我不需要将menu置于do-while循环中。当然，这意味着只要用户名和密码不正确，它就不会到达该部分。如何退出是你的练习。

截图：

enter image description here

希望这有帮助。

如何留下循环

3 个答案: