如何使输出只打印一次

Question

我正在做一个猜数字游戏。问题是：当数字被猜对并且计算机询问用户是否想再次玩时，它会接受输入并在退出游戏或重新启动之前再次询问相同的问题。

我尝试通过添加来调整 shouldPlayAgain() 函数

else if(input == 'N') 
   return false; 

但我仍然遇到同样的问题。这是我的代码： *注意 int main 函数必须是这样的。我无法对其添加更改。

#include <iostream>
using namespace std;

//Function prototypes
void playOneGame();
char getUserResponseToGuess(int);
int getMidpoint(int, int);
bool shouldPlayAgain();

int main()
{
    do
    {
        playOneGame();
    } while (shouldPlayAgain());

    return 0;
}

void playOneGame()
{
    int low = 1;
    int high = 100;
    int guess = getMidpoint(low, high);
    char response;

    cout << "Welcome! Please think of a number from 1 to 100.\n";
    response = getUserResponseToGuess(guess);

    //Keeps guessing until it guesses the correct number
    while(response != 'C')
    {
        if(response == 'H')
        {
            low = guess+1;
            guess = getMidpoint(low, high);
            response = getUserResponseToGuess(guess);
        }
        else if(response == 'L')
        {
            high = guess-1;
            guess = getMidpoint(low, high);
            response = getUserResponseToGuess(guess);
        }
    }

    if(response == 'C')
    {
        shouldPlayAgain();
    }
}

char getUserResponseToGuess(int guess)
{
        char HLC;
        cout << "Is the number " << guess << " ? (H/L/C)\n";
        cin >> HLC;
        return HLC;
}

int getMidpoint(int low, int high)
{
    int mid = (low+high)/2;
    return mid;
}

bool shouldPlayAgain()
{
    char input;
    cout << "Would you like to play again? (Y/N)\n";
    cin >> input;

    if(input == 'Y')
    {
        return true;
    }
    else return false;
}

这是一个示例输出。我不明白为什么它在最终执行之前两次打印问题。任何帮助表示赞赏。谢谢。

Welcome! Please think of a number from 1 to 100.
Is the number 50 ? (H/L/C)
L
Is the number 25 ? (H/L/C)
H
Is the number 37 ? (H/L/C)
H
Is the number 43 ? (H/L/C)
L
Is the number 40 ? (H/L/C)
C
Would you like to play again? (Y/N)
Y
Would you like to play again? (Y/N)
Y
Welcome! Please think of a number from 1 to 100.
Is the number 50 ? (H/L/C)

Answer 1

您应该从playOneGame() 中删除对shouldPlayAgain() 的调用。

从playOneGame() 每次返回后，都会在main() 中调用它。