一個班級成員的相同論點：簡化？

Question

我想知道如果你有C++類中的成員具有相同的參數，如果有一種方法來簡化這個？例如：

#include <iostream> 
using namespace std: 

class bankAccount 
{ 
public: 
     bankAccount() 
     { 
      privAcct = "MyAccount"; 
      privPin = "MyPin"; 
     { 
     void changeBalance(string acct, string pin) 
     { 
      if(acct == privAcct && pin == privPin) 
      { 
       cout << "YAY! You can do this!" << endl; 
      } 
     } 
     void otherMethod(string acct, string pin) 
     { 
      if(acct == privAcct && pin == privPin) 
      { 
       cout << "YAY! You can do this!" << endl; 
      } 
     } 
private: 
     string privAcct, privPin; 
}; 

正如你可以看到他們都採取同樣的參數，並都需要在相同的條件是真實的訪問方法的「肉」。

雖然我在想可能會創建一個方法，然後爲了訪問方法的不同部分而使用switch語句，但我希望能夠訪問「changeBalance」部分或「 otherMethod「類的一部分。我只是不確定是否有辦法讓這個「整潔」一點，或簡化它？

我的整個代碼是這樣的：

#include <cstdlib> 
#include <iostream> 
#include <string> 

using namespace std; 
void accountInfo(); 

class account{ 
public: 
    //constructor 
    account() 
    { 
     balance = 0.00; 
     accountNum = "0303"; 
     pin = "2222"; 
    } 
    //default for wrong entry response 
    void wrongEntry() 
    { 
     cout << "Wrong account number or PIN. Please try again.\n" << endl; 
    } 
    //change pin number 
    void changePin(string actNum, string oldPin) 
    { 
     if(actNum == accountNum && oldPin == pin) 
     { 
      string newPin; 
      cout << "Please enter in a new pin: "; 
      cin >> newPin; 
      pin = newPin; 
      cout << "Thank you." << endl; 
     } 
     else 
     { 
      wrongEntry(); 
     } 
    } 
    //change balance 
    void changeBalance(string actNum, string pinnum) 
    { 
     if(actNum == accountNum && pinnum == pin) 
     { 
      double newAdd; 
      cout << "Your current balance is " << balance << "\nPlease enter the additional amount you would like to deposit: "; 
      cin >> newAdd; 
      balance += newAdd; 
      cout << "Your new balance is " << balance << ".\nThank you.\n" << endl; 

     } 
     else 
     { 
      wrongEntry(); 
     } 
    } 
    //print balance and account # 
    void printBalance(string actNum, string pinnum) 
    { 
     if(actNum == accountNum && pinnum == pin) 
     { 
      cout << "For account #"<< accountNum << endl; 
      cout << "Your current balance is $" << balance << ".\nThank you.\n"; 
     } 
     else 
     { 
      wrongEntry(); 
     } 
    } 

private: 
    string accountNum, pin; 
    double balance; 
}; 

int main() 
{ 
    int input; 
    string aN, pin; 
    account bo; 

    while(1) 
    { 
     cout << "Please enter account number: "; 
     cin >> aN; 
     cout << "Please enter corresponding PIN: "; 
     cin >> pin; 

     ///options 
     cout << "Please choose from the following options:\n"; 
     cout << " 1. View Balance\n 2. Deposit\n 3. Change pin\nChoice: "; 
     cin >> input; 
     cout << endl; 
     switch(input) 
     { 
     case 1: 
      bo.printBalance(aN, pin); 
      break; 
     case 2: 
      bo.changeBalance(aN, pin); 
      break; 
     case 3: 
      bo.changePin(aN, pin); 
      break; 
     default: 
      cout << "The information you entered does not seem to match our records.\nPlease try again.\n"; 
      break; 
     } 
     char response; 
     cout << "\nAre you done with your transaction?: Y/N"; 
     cin >> response; 
     cout << endl; 
     if(response == 'Y' || response == 'y') 
     { 
      return 0; 
     } 
    } 
} 

我知道我應該把我的班級在不同的頭文件，以聲明的構造正確一起，但我只是希望簡化代碼。

謝謝。

Answer 1

您可以將「受限制」接口封裝爲另一種類型，並且只允許使用有效憑證創建該類型。喜歡的東西

class Account { 
public: 
    struct Accessor { 
     Accessor(Account & a, string acct, string pin) : account(&a) { 
      if (acct != a.privAcct || pin != a.privPin) { 
       throw BadCredentials(); 
      } 
     } 

     void changePin() {account->changePin();} 
     // and other functions 

     Account * account; 
    }; 
private: 
    string privAcct, privPin; 

    // Only accessible through an Accessor 
    void changePin(); 
    // and other functions 
}; 

用法：

Accessor(account, acct, pin).changePin(); 

或者，如果整個界面限制，您可以簡單地把驗證在Account本身的構造。

Answer 2

你可以考慮把它當做交易。

（僞代碼/設計）

transactionValid = BeginTransation(account, pin); 
if (transactionValid) 
{ 
    changePin(...); 
    chackBalance(...); 
} 
EndTransaction(); 

Answer 3

您可以創建一個新的類LoginCredentials其中包含acct和pin爲私有成員和成員函數，將檢查是否相等。