Question

我已经尝试过所有可以让它发挥作用的东西。我已经研究了这么多，并尝试了我能找到的所有可能的解决方案。我在Q ++中使用QtCreator和编码。

int main() {
    int Die1, Die2;
    cout << "Die 1: ";
    cin >> Die1;
    cout << "Die 2: ";
    cin >> Die2;
    while ( (Die1+Die2 != '2')||(Die1+Die2 != '3')||(Die1+Die2 != '5')||(Die1+Die2 != '7')||(Die1+Die2 != '9')||(Die1+Die2 != '11')) {
        cout << "You Rolled a " <<Die1<< " and " <<Die2<< " for a Total of " <<Die1 + Die2<<". Please Roll Again\n";
        cout << "Die 1: ";
        cin >> Die1;
        cout << "Die 2: ";
        cin >> Die2;
    }

Answer 1

您的代码：

(Die1+Die2 != '2') ||
(Die1+Die2 != '3') ||
(Die1+Die2 != '5') ||
(Die1+Die2 != '7') ||
(Die1+Die2 != '9') ||
(Die1+Die2 != '11')

将评估为：

A.
( 1+1 != 2 ) = false
( 1+1 != 3 ) = true
break;

B.
( 1+2 != 2 ) = true
break;

C.
( 1+3 != 2 ) = true
break;

...
( 1 + ( x>2 ) != 2 ) = true

这里的逻辑本质上存在缺陷，因为它总是评估为 true 。

修改的

正如其他人所指出的那样，您也在尝试将这些整数与其char值进行比较，这些值与您期望的值不符。

'0' = (hex)0x30 = (int)48
'1' = (hex)0x31 = (int)49
'2' = (hex)0x32 = (int)50
...

要解决此问题，您应该与实际数字进行比较：

( Die1 + Die2 != 2 )

Answer 2

你可能想检查一下你的状况！

由于条件是否定的，你可以使用AND（＆amp;＆amp;）而不是OR（||）来组合它们。

修改正如所指出的，你应该将Die1和Die2与数字进行比较。如果您想保留字符，可以添加字符＆＃39; 0＆＃39; 0到Die1和Die2的总和。

这似乎工作正常：

#include <bits/stdc++.h> using namespace std; int main() { int Die1, Die2; cout << "Die 1: "; cin >> Die1; cout << "Die 2: "; cin >> Die2; while ( (Die1+Die2 != 2)&&(Die1+Die2 != 3)&&(Die1+Die2 != 5)&&(Die1+Die2 != 7)&&(Die1+Die2 != 9)&&(Die1+Die2 != 11)) { cout << "You Rolled a " <<Die1<< " and " <<Die2<< " for a Total of " <<Die1 + Die2<<". Please Roll Again\n"; cout << "Die 1: "; cin >> Die1; cout << "Die 2: "; cin >> Die2; }

}

while循环运行时为false

2 个答案: