不確定為什么代碼不起作用 C++ simple

Question

試圖阻止用戶輸入字符。 這段代碼在我的腦海中很有意義。 我所做的第一個 if 語句按預期工作（它阻止用戶輸入字符）。 但是當用戶做出正確選擇時，開關直接進入默認情況。 在我輸入錯誤處理 if 語句之前，開關工作正常。 為幫助歡呼

void Input()
{
char errorhandle;
int a;
cout << "It's " << player << "'s turn Enter where you want your shape: ";
cin >> errorhandle;

if (errorhandle < '0' || errorhandle > '9')
{
    cout << "You have not entered a number try again!" << endl;
    Input();
}
else
{
    a = (int)errorhandle;
}

switch (a)
{
case 1:
    if (board[0][0] == '1')
    {
        board[0][0] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 2:
    if (board[0][1] == '2')
    {
        board[0][1] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 3:
    if (board[0][2] == '3')
    {
        board[0][2] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 4:
    if (board[1][0] == '4')
    {
        board[1][0] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 5:
    if (board[1][1] == '5')
    {
        board[1][1] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 6:
    if (board[1][2] == '6')
    {
        board[1][2] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 7:
    if (board[2][0] == '7')
    {
        board[2][0] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 8:
    if (board[2][1] == '8')
    {
        board[2][1] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

case 9:
    if (board[2][2] == '9')
    {
        board[2][2] = player;
    }
    else
    {
        cout << "The place is already in use, try again!" << endl;
        Input();
    };
    break;

default:
    cout << "You have entered an invalid option, try again" << endl;
    Input();
}

}

Answer 1

問題是，當您確定錯誤時，您再次調用您的函數：

Input();

當用戶然后輸入一個好的數字時，它會使用好的輸入執行切換。 然后它返回給調用者，在錯誤處理后恢復，並使用未初始化a a 第二次執行 switch

還有另一個問題：當您使用a = (int)errorhandle;將輸入轉換為整數時a = (int)errorhandle; , '1' 的輸入將被轉換為 '1' 的 ascii 值而不是 1。所以你的 case 值應該堅持引用的值。

潛在修正：

while ( (cin >> errorhandle) && (errorhandle < '0' || errorhandle > '9') )
    cout << "You have not entered a number try again! " << endl;
a = errorhandle-'0';
switch (a)
...

Answer 2

在這一行：

a = (int)errorhandle;

您正在將 ascii char轉換為int 。 '1' 的值和 1 不一樣。看ascii 表。

同樣在您遞歸調用Input()您將繼續在未初始化的switch語句中使用a 。

if (errorhandle < '0' || errorhandle > '9') {
    cout << "You have not entered a number try again!" << endl;
    Input();
    return; // Stop execution after this line. 
            // This should be done in all cases of a call to input. 
} else {
    a = (int)(errorhandle - '0');
}

Answer 3

以前的答案解釋了發生了什么，您可以像這樣修復它：

if (errorhandle < '0' || errorhandle > '9')
{
    cout << "You have not entered a number try again!" << endl;
    Input();
    return; // < new | stops the function
}

在這種情況下我會避免遞歸，但這也有效。

並且您不能像那樣將 char 轉換為 int。 甚至不要轉換為 int 只是比較 switch 語句中的 char 值。