如果不使用循环，我如何迭代代码指令

时间：2017-01-15 03:32:13

标签： c++ iteration restart

当用户输入错误的密码时，代码将在他们尝试两次代码停止时重新启动，但是在我的代码中，当用户输入错误的信息并且它通过了password_attempts时，它继续执行其他一切似乎是工作得很好

#include <iostream>
#include <string>

using namespace std;

int main()
{

    string myusername;
    string mypassword;
    bool Access_granted;
    int password_attempts;
    cout << "Enter your username: ";
    cin >> myusername;
    cout << "Enter your password: ";
    cin >> mypassword;
    if (myusername == "veasy62" && mypassword == "a65908") {
        Access_granted = true;
        cout << "Access granted veasy62\n";
    }
    else if (myusername == "tveasy62" || mypassword == "a1065908") {
        Access_granted = true;
        cout << "Access granted tveasy62\n";
    }
    else {
        Access_granted = false;
        cout << password_attempts;
        cout << "Access Denied, Sorry try again\n";
        if (Access_granted == false) {
            if (password_attempts = 2) {
                password_attempts = password_attempts + 1;
                return main();
            }
            else {
                cout << "Sorry you have ran out of attempts\n";
            }
        }
    }
}

6 个答案:

答案 0 :(得分：2)

你的代码得到了Litttle buggy。 而不是＆amp;＆amp;你用过了|| 在条件允许的情况下，= = 2的insteas = = password_attempts = 2 如果（错误检查条件）不必要 我删除了所有这些，这是您想要实现的任务的代码

#include <iostream>
#include <string>

using namespace std;

int main()
{

    string myusername;
    string mypassword;
    bool Access_granted;
    int password_attempts=0;

    HERE : cout << "Enter your username: ";
    cin >> myusername;
    cout << "Enter your password: ";
    cin >> mypassword;
    if (myusername == "veasy62" && mypassword == "a65908") {
        Access_granted = true;
        cout << "Access granted veasy62\n";
        break;
    }
    else if (myusername == "tveasy62" && mypassword == "a1065908") {
        Access_granted = true;
        cout << "Access granted tveasy62\n";
        break;
      }
    else {
        password_attempts = password_attempts + 1;
        Access_granted = false;
        cout << password_attempts;
        cout << "Access Denied, Sorry try again\n";

            if (password_attempts == 2) {
               cout << "Sorry you have ran out of attempts\n";
               return 0;

              //to exit main().can also use exit(0) using additional libraries
           }
          else{
           goto HERE;
        }
        }

 }

答案 1 :(得分：1)

如果你不想使用循环，你可以使用goto而不是返回。你可以使用“endl”而不是“\ n”。你可以试试这个 - 

#include <iostream>
#include <string>
using namespace std;

int main()
{
   string myusername;
   string mypassword;
   bool Access_granted;
   int password_attempts=0;
   restart:
   cout << "Enter your username: ";
   cin >> myusername;
   cout << "Enter your password: ";
   cin >> mypassword;
   if(myusername == "veasy62" && mypassword == "12")
   {
       Access_granted = true;
       cout<<"Access granted veasy62" << endl;
   }
   else if (myusername == "tveasy62" || mypassword == "a1065908" )
   {
      Access_granted = true;
      cout<<"Access granted tveasy62" << endl;
   }
   else
   {
      Access_granted = false; 


       if(Access_granted == false)
      {
          if(password_attempts < 2 )
          {
              cout << "Access Denied, Sorry try again" << endl;
              password_attempts = password_attempts + 1;
              cout << "you have remaining " << 2-password_attempts << " time   chances after this attempt" << endl;
              goto restart; 
          }
          else
          {
              cout << "Sorry you have ran out of attempts\n";
          }
      }
  }

   return 0;
}

答案 2 :(得分：1)

所以你说你不能使用循环。那么这是一个使用递归模板的解决方案。没有＆＃34; gotos＆＃34;，可以通过将模板参数更改为更大的数字来轻松调整。

#include <string>
#include <iostream>

template <int attempts>
struct password_entry
{
    static bool get_entry(std::string& myusername, std::string& mypassword)
    {
        std::cout << "Enter your username: ";
        std::cin >> myusername;
        std::cout << "Enter your password: ";
        std::cin >> mypassword;
        if (myusername == "veasy62" && mypassword == "12")
        {
            std::cout << "Access granted veasy62" << std::endl;
            return true;
        }
        if ( attempts > 1 )
            std::cout << "Access denied.  You have " << attempts - 1 << " attempts remaining\n";
        return password_entry<attempts - 1>::get_entry(myusername, mypassword);
    }
};

template <> 
struct password_entry<0>
{
    static bool get_entry(const std::string&, const std::string&)
    {
        std::cout << "Sorry you have ran out of attempts\n";
        return false;
    }
};


int main()
{
    std::string mypass, myuser;

    // max 2 attempts 
    bool access_granted = password_entry<2>::get_entry(myuser, mypass);
    if ( access_granted )
       std::cout << "Welcome current user";
    else
       std::cout << "Please call 555-5555 to reset your password";
}

要调整尝试次数，只需更改password_entry模板参数调用。

Live Example

编辑：调整后的代码不使用构造函数。

答案 3 :(得分：0)

所以我从你那里得知你想让用户输入他的密码两次，如果两者都错了，那么完成程序吧？ 那么你可以这样写。

编辑：）

#include <iostream>
#include <string>

using namespace std;

int main()
{

    string myusername;
    string mypassword;
    bool Access_granted;
    int password_attempts=0;




        cout << "Enter your username: ";
        cin >> myusername;
        cout << "Enter your password: ";
        cin >> mypassword;
        if (myusername == "veasy62" && mypassword == "a65908") {
            Access_granted = true;
            cout << "Access granted veasy62\n";
        }
        else if (myusername == "tveasy62" || mypassword == "a1065908") {
            Access_granted = true;
            cout << "Access granted tveasy62\n";
            system("pause");
            return 0;
        }
        else {
            Access_granted = false;
            cout << "Access Denied, Sorry try again\n";
            cout << "Enter your username: ";
            cin >> myusername;
            cout << "Enter your password: ";
            cin >> mypassword;
            if (myusername == "veasy62" && mypassword == "a65908") {
                Access_granted = true;
                cout << "Access granted veasy62\n";
            }
            else if (myusername == "tveasy62" || mypassword == "a1065908") {
                Access_granted = true;
                cout << "Access granted tveasy62\n";
                system("pause");
                return 0;
            }
            else
            {


                    cout << "Sorry you have ran out of attempts\n";
                    cout << "Access Denied";
                    system("pause");
                    return -1;
                }
            }



    return 0;
}

答案 4 :(得分：0)

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string myusername;
    string mypassword;
    bool Access_granted;
    int password_attempts=0;
    cout << "Enter your username: ";
    cin >> myusername;
    cout << "Enter your password: ";
    cin >> mypassword;
    if (myusername == "veasy62" && mypassword == "a65908") {
        Access_granted = true;
        cout << "Access granted veasy62\n";
    }
    else if (myusername == "tveasy62" || mypassword == "a1065908") {
        Access_granted = true;
        cout << "Access granted tveasy62\n";
    }
    else {
        Access_granted = false;
        cout << password_attempts;
        cout << "Access Denied, Sorry try again\n";
        if (Access_granted == false) {
            if (password_attempts <=2) {
                password_attempts = password_attempts + 1;
                return main();
            }
            else {
                cout << "Sorry you have ran out of attempts\n";
            }
        }
    }
}

答案 5 :(得分：0)

使用转到 或者，如果您喜欢模块，请尝试使用递归函数。它将像您尝试的那样工作。

public int accessAllowed(int notrys=0)
{
if(notrys<2)
return 0;

string username,password;

//take username password

If(//condition 1)
return 1;

else if(//condition 2)
return 1;

else
return accessAllowed(notrys+1);
}

虽然goto会很有效率。

您的代码中还有很多错误

  1. Nooftrys未初始化（我认为您需要0作为初始值）
  2. ||取代＆amp;＆amp;
    3. 检查trys溢出时
  3. = 2，使用＆lt; 2进行2次尝试。
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