获取一个字符串在密码功能中工作 [英] getting a string to work in password function

问题描述

很抱歉我的可怜的解释我很新的这个。我目前正在开发一个密码功能，但我有问题。我设置帐户名称为字符串john和帐户密码为int 1111.密码工作正常，但字符串导致错误。当我把const string name =john改成一个随机整数代码工作正常。

Hi apologies for my poor explanation I'm very new to this. I am currently working on a password function but I'm having problems. I have set the account name to string "john" and the account password to int 1111. The password works fine but the string is causing the error. When I change "const string name = "john"" into a random integer the code works fine.

我希望有人能找到我错误的地方。

I was hoping someone could spot where I'm going wrong?

bool login() {

   const int password = 1111; 
   int passwordAttempt;
   const string name = "john";
   string nameAttempt;
   int attempts = 0;

   std::cout << "Please Enter Password:" << std::endl; //first attempt at account number & password
   std::cin >> passwordAttempt;
   std::cout << "Enter Name:"<< std::endl;
   std::cin >> nameAttempt;

if (passwordAttempt == password && nameAttempt == name) {
   return true;
   } 

else 

while (passwordAttempt!=password || nameAttempt!=name) {
     if(attempts++ ==2)//.. a loop for two more attempts
     {
       std::cout << "You have entered the wrong details three times. The application will be terminated now" << std::endl;
       return false;
       break;
     }
       std::cout<<"Incorrect password. Try again"<<std::endl;
       std::cout<< "" <<std::endl;
       std::cout << "Please Enter Password:" << std::endl;
       std::cin >> passwordAttempt;
       std::cout << "Enter Name:"<< std::endl;
       std::cin >>nameAttempt;
     }

     }




using namespace std;

int main() {

bool loggedIn = login();

if (loggedIn){
cout << "Logged In" << endl;
}

else if (!loggedIn){
cout << "Not Logged" << endl;
}

system("pause");
return 0;
}

推荐答案

1. 将 std :: 添加到您的字符串声明（因此您将拥有 std :: string ）。


2. 在while循环结束后，函数 login 没有任何值返回。添加 return true 。也总是建立警告并启用！

1. Add std:: to your string declarations (so you will have std::string).
2. After while loop there's end of function login without any value return. Add return true there. Also always build with warnings enabled!

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