使用 getline() 将字符串输入程序会导致某种溢出到下一个输入

Question

我的程序中有很多问题都与输入有关。在程序中，我要求用户输入的第一件事是他们的名字，我用这个

cout << "Please tell me your name." << endl;
getline(cin, user_name);
cout << "Hello " << user_name << " and welcome to Fantasy Battle!" << endl;

其中 user_name 在别处声明为字符串变量。这部分似乎没有问题，因为以下消息正确显示在屏幕上

用户的下一个输入来自此代码

{
cout << "Hello, what would you like to do?" << endl;
cout << "1. Play" << endl << "2. Exit" << endl;
cout << "Please enter the number corresponding to your choice from the list 
above." << endl;
for(;;)
{
    if(cin >> menuChoice)
    {
        if(cin.get() == '.')
        {
            cin.clear();
            cin.ignore(10000, '\n');
        }
        if(menuChoice == 1 || menuChoice == 2)
            break;
        else
        {
            cout << "You did not enter a valid menu option. Please try 
            again." << endl;
            cin.clear();
            cin.ignore(100000, '\n');
        }
    }
    else
    {
        cout << "You did not enter a valid menu option. Please try again." 
        << endl;
        cin.clear();
        cin.ignore(100000, '\n');
    }
}
if(menuChoice == 2)
{
    return 2;
}
else
{
    //setup a fight code further down
}

第一个运行如果我输入 2 它将从 main 成功退出程序，或者如果输入 1 它将运行通过战斗功能。但是，如果我经历了 1 次战斗并返回要求我输入 1 或 2 进行播放或退出的程序，我可以无限次输入 2 并且它不会退出程序。我不确定是什么原因造成的。

for(;;)
    {
        game = menu();
        if(game == 2)
        {
            break;
        }
        else
        {
            fight();
        }
     }
     return 0;

我程序的 menu() 函数内的代码如下，其中包含我程序的其余输入。我正在使用 getline(cin, fighterName) 从用户那里获取一个字符串，用作他们想要创建的每个角色的名称我遇到的问题是它开始只是在不询问的情况下将字符名称保存为空。

cout << "How many fighters should be on Team 1?" << endl;
//Input Validation
for(;;)
{
    if(cin.get() == '.')
    {
        cin.clear();
        cin.ignore(100000, '\n');
    }
    if(cin >> team1Size)
    {
        if(team1Size <= 0)
        {
            cout << "The team must be a size of at least 1 fighter. Please try again." << endl;
            cin.clear();
            cin.ignore(100000, '\n');
        }
        else
        {
            break;
        }
    }
    else
    {
        cout << "You did not enter a valid number. Please try again." << endl;
           cin.clear();
            cin.ignore(100000, '\n');
    }
}

cout << "How many fighters should be on Team 2?" << endl;
//Input Validation
for(;;)
{
    if(cin.get() == '.')
    {
        cin.clear();
        cin.ignore(100000, '\n');
    }
    if(cin >> team2Size)
    {
        if(team2Size <= 0)
        {
            cout << "The team must be a size of at least 1 fighter. Please try again." << endl;
            cin.clear();
            cin.ignore(100000, '\n');
        }
        else
        {
            break;
        }
    }
    else
    {
        cout << "You did not enter a valid number. Please try again." << endl;
           cin.clear();
            cin.ignore(100000, '\n');
    }
}

//Set up team 1 
cout << "Begin setup for team 1:" << endl << endl;
for(int i = 0; i < team1Size; i++)
{
    cout << "Which character type should fighter " << i+1 << " be?" << endl;
    cout << "1. Barbarian" << endl;
    cout << "2. BlueMen" << endl;
    cout << "3. Vampire" << endl;
    cout << "4. Medusa" << endl;    
    cout << "5. Harry Potter" << endl;
    cout << "Please enter the number corresponding to your choice from the list above." << endl;
    for(;;)
    {
        if(cin.get() == '.')
        {
            cin.clear();
            cin.ignore(100000, '\n');
        }
        if(cin >> fighterType)
        {
            if(fighterType < 1 || fighterType > 5)
            {
                cout << "You did not enter a valid choice. Please try again." << endl;
                cin.clear();
                cin.ignore(100000, '\n');
            }
            else
                break;
        }
        else
        {
            cout << "You did not enter a valid choice. Please try again." << endl;
            cin.clear();
            cin.ignore(100000, '\n');
        }
    }
    //Now that we have the desired type of the fighter we must add a fighter of the correct type to the linked list
    //representing team 1. We will do so by calling the add function of the linked list
    cout << "Please enter the name of this fighter." << endl;
    getline(cin, fighterName);
    if(fighterType == 1)
    {
        team1.addBack("Barbarian", fighterName);
    }
    else if(fighterType == 2)
    {
        team1.addBack("BlueMen", fighterName);
    }
    else if(fighterType == 3)
    {
        team1.addBack("Vampire", fighterName);
    }
    else if(fighterType == 4)
    {
        team1.addBack("Medusa", fighterName);
    }
    else
    {
        team1.addBack("HarryPotter", fighterName);
    }

}

cout << "Team 1 has been created!" << endl << endl;

//Set up team 2 
cout << "Begin setup for team 2:" << endl << endl;
for(int i = 0; i < team2Size; i++)
{
    cout << "Which character type should fighter " << i+1 << " be?" << endl;
    cout << "1. Barbarian" << endl;
    cout << "2. BlueMen" << endl;
    cout << "3. Vampire" << endl;
    cout << "4. Medusa" << endl;    
    cout << "5. Harry Potter" << endl;
    cout << "Please enter the number corresponding to your choice from the list above." << endl;
    for(;;)
    {
        if(cin.get() == '.')
        {
            cin.clear();
            cin.ignore(100000, '\n');
        }
        if(cin >> fighterType)
        {
            if(fighterType < 1 || fighterType > 5)
            {
                cout << "You did not enter a valid choice. Please try again." << endl;
                cin.clear();
                cin.ignore(100000, '\n');
            }
            else
                break;
        }
        else
        {
            cout << "You did not enter a valid choice. Please try again." << endl;
            cin.clear();
            cin.ignore(100000, '\n');
        }
    }
    //Now that we have the desired type of the fighter we must add a fighter of the correct type to the linked list
    //representing team 2. We will do so by calling the add function of the linked list
    cout << "Please enter the name of this fighter." << endl;
    getline(cin, fighterName);
    if(fighterType == 1)
    {
        team2.addBack("Barbarian", fighterName);
    }
    else if(fighterType == 2)
    {
        team2.addBack("BlueMen", fighterName);
    }
    else if(fighterType == 3)
    {
        team2.addBack("Vampire", fighterName);
    }
    else if(fighterType == 4)
    {
        team2.addBack("Medusa", fighterName);
    }
    else
    {
        team2.addBack("HarryPotter", fighterName);
    }
}


cout << "Team 2 has been created!" << endl << endl;

cout << "Let the fight begin!" << endl << endl;

return 0;

}

我的代码的最后一段输入如下，它只是要求用户输入一个字符 y 或 n，然后在输入 y 时执行一个函数。

cout << "Would you like to see the contents of the loserPile?" << endl;
    cout << "Please enter y for yes or n for no" << endl;
    for(;;)
    {
        if(cin >> displayLosers)
        {
            if(displayLosers != 'y' && displayLosers != 'n')
            {
                cout << "You did not enter either y or n. Please try again." << endl;
                cin.clear();
                cin.ignore(100000, '\n');
            }
            else
                break;
        }
        else
        {
            cout << "You did not enter either y or n. Please try again." << endl;
            cin.clear();
            cin.ignore(100000, '\n');
        }
    }
    if(displayLosers == 'y')
    {
        losers.displayPile();
    }

如果有人能指出我在获取用户输入时出错的地方，我将不胜感激，因为我运行正在尝试我所知道的事情。

Answer 1

您添加 if(cin.get() == '.')

会造成很多问题

>> 运算符会将输入字符串 "1." 转换为 1，如果您调用 ignore(...,'\n') 则 . 和 [ 之前的任何其他字符=18=] 将被忽略。也没有必要测试 if(cin >> number){...}。您可以将值初始化为 -1 以指示错误：

int menuChoice;
for(;;)
{
    menuChoice = -1;
    cin >> menuChoice;
    cout << menuChoice;
    cin.clear();
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    if(menuChoice == 1 || menuChoice == 2)
    {
        cout << menuChoice << "\n";
        break;
    }
    cout << "You did not enter a valid menu option. Please try again." << endl;
}

只要确保您使用的是正确的输入即可。对于是或否选项，输入应为 char:

cout << "enter y or n\n";
for(;;)
{
    char val;
    cin >> val;
    if(val != 'y' && val != 'n')
    {
        cout << "You did not enter either y or n. Please try again." << endl;
        cin.clear();
        cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    }
    else
        break;
}

使用 getline() 将字符串输入程序会导致某种溢出到下一个输入

using getline() to input strings into the program is causing some kind of overflow into the next input

c++

validation

menu

input

getline