为什么我的cout打印两次？

Question

I'm working on a text-based game and want the player to choose single or multiplayer. I have the following check in my code:

cout << "Welcome to Hangman!\nIs this a [s]ingle or [m]ultiplayer game?\nGame mode: ";

int type = cin.get();
cin.clear();

//While input isn't single character s or m (either case)
while (type != 115 && type != 109 && type != 83 && type != 77) {
    cout << endl << "Please try again.\nGame mode: ";
    cin.clear();
    type = cin.get();
}

What's happening is if the player gives an invalid input, "Please try again. Game mode: " prints twice, but if the player just hits enter it prints once. I'm a beginner to C++ and read that cin.clear() sometimes fixes this issue, but nothing has worked so far.

Answer 1

You should call cin.ignore(std::numeric_limits<std::streamsize>::max(), '\\n'); after cin.clear()

cout << "Welcome to Hangman!\nIs this a [s]ingle or [m]ultiplayer game?\nGame mode: ";

int type = cin.get();
cin.clear();

//While input isn't single character s or m (either case)
while (type != 115 && type != 109 && type != 83 && type != 77) {
    cout << endl << "Please try again.\nGame mode: ";
    cin.clear();
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    type = cin.get();
}

Edit: as dyp mentioned in comments sync is not guaranteed to do anything useful in this case (its behaviour is implementation defined) so the only portable way is to discard characters until new line