返回 function 的 output 作為 main() 的返回碼

Question

我在這里自學，所以我沒有可用的 model 答案。

處理程序流示例並嘗試基於二進制搜索獲得數字猜測器。 我已經讓它成功運行並捕獲邊緣情況，但一個目標是讓main()返回所做的猜測次數。 我將主代碼重構為單獨的 function 以使其更清晰，但我無法正確獲取返回代碼，我懷疑它與變量 scope 有關，但無法弄清楚。

#include <iostream>

using namespace std;

int guessNumber(int highest, int lowest, int lAttempts)
{
    int guess = lowest + ((highest - lowest) * 0.5);
    char response = 'a';

    lAttempts++;
    cout << "My guess is " << guess << ", am I correct?" << endl;
    cout << "(y)es/too (h)igh/too (l)ow/(q)uit" << endl;
    cin >> response;

    while (response != 'y' && response != 'h' && response != 'l' && response != 'q')
    {
        cout << "I'm sorry, I didn't understand that" << endl;
        cout << "(y)es/too (h)igh/too (l)ow/(q)uit" << endl;
        cin >> response;
    }

    switch (response)
    {
    case 'y':
        cout << "I guessed correctly after " << lAttempts << " attempts";
        break;
    case 'h':
        highest = guess;
        guessNumber(highest, lowest, lAttempts);
        break;
    case 'l':
        lowest = guess;
        guessNumber(highest, lowest, lAttempts);
        break;
    case 'q':
        cout << "Exiting program";
        break;
    }
    return lAttempts;
}

int main()
{
    cout << "Think of a number between 1-100" << endl;

    int highest = 100;
    int lowest = 0;
    int attempts = 0;

    attempts = attempts + guessNumber(highest, lowest, attempts);
    return attempts;
}

cout返回正確的嘗試次數，但程序（因此main() ）始終以 1 退出。

我在這里想念什么？

謝謝。

Answer 1

您缺少在 switch 語句中更新您的嘗試變量。
應該是這樣的。

lAttempts = guessNumber(highest, lowest, lAttempts);