比較函數調用返回的字符串

Question

我似乎無法弄清楚爲什麼get_password函數調用總是會返回qwert，無論我傳入函數的字符串是什麼。我的問題是我無法看到這個函數中字符串比較出了什麼問題。

string get(string askfor, int numchars, string input) 
{ 
    cout << askfor << "(" << numchars << " characters): "; 
    cin >> input; 
    return input; 
} 

string get_password(string name) 
{ 
    string pwd; 
    if (name == "botting"){ 
     pwd = "123456"; 
    } 
    else if (name == "ernesto") { 
     pwd = "765432"; 
    } 
    else if (name == "tong") { 
     pwd = "234567"; 
    } 
    else { 
     pwd = "qwert"; 
    } 

    return pwd; 
} 

int main() 
{ 
    string name; 
    string pwd; 
    string passwd; 
    cout << "Address of name =" << &name << "\n"; 
    cout << "Address of pwd =" << &pwd << "\n"; 
    cout << "Address of passwd =" << &passwd << "\n"; 

    bool authenticated = false; 
    while (!authenticated) 
    { 
     // call one 
     string name1 = get("Name", 7, name); 
     cout << "call one returned: " << name1 << endl; 

     // call two 
     string pass1 = get_password(name); 
     cout << "call two returned: " << pass1 << endl; 

     //call three 
     string pass2 = get("Password", 7, passwd); 
     cout << "call three returned: " << pass2 << endl; 

     // compare the two passwords 
     authenticated = false; 
     if (pass1 == pass2) { 
      cout << "Welcome " << name << "\n"; 
      authenticated = true; 
     } 
     else { 
      cout << "Please try again\n"; 
     } 
    } 

    return 0; 
} 

Answer 1

通name1第二個呼叫：

// call one 
    string name1 = get("Name", 7, name); 
    cout << "call one returned: " << name1 << endl; 

    // call two 
    string pass1 = get_password(name1); // change this variable 
    cout << "call two returned: " << pass1 << endl; 

的get()返回名稱name1字符串，並且不更新name變量本身，因爲這name仍然空字符串。

Answer 2

那是因爲你從來沒有分配任何東西name。它被初始化爲""，並且永遠不會改變。由於這與get_password中的任何案例都不匹配，因此它總是屬於else的情況，這將產生"qwert"。

我認爲這個問題是，你的get功能將得到更好的這樣寫：

string get(string askfor, int numchars) 
{ 
    string input; // input shouldn't be an argument 
    cout << askfor<<"("<<numchars<<" characters): "; 
    cin >> input; 
    return input; 
} 

而且你可以使用結果是分配到name：

name = get("Name", 7); 

Answer 3

你只需要通過參考而不是值。

string get(string askfor, int numchars, string &input) 
{ 
    cout << askfor << "(" << numchars << " characters): "; 
    cin >> input; 
    return input; 
}