比较函数调用返回的字符串

Question

我似乎无法弄清楚为什么 get_password 函数调用总是 return qwert 无论我将什么字符串传递给函数。我的问题是我看不出这个函数中的字符串比较出了什么问题。

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

那是因为您从未分配 name 任何东西。它被初始化为 ""，并且永远不会改变。由于匹配 get_password 中的 none 个案例，它总是落入 else 个案例，从而产生 "qwert"。

我认为问题是您的 get 函数最好这样写：

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

并且您可以使用它的结果分配给 name:

name = get("Name", 7);

Answer 2

您只需要通过 reference 而不是 value。

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

Answer 3

将name1传递给第二个调用：

    // 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;

get() returns 名称为 name1 字符串并且不会更新 name 变量本身，因此 name 仍然是空字符串。

比较函数调用返回的字符串

compare strings returned from function calls

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string