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Question

为什么这段代码有两个不同的输出（GCC 4.5.1）（我评论过重要的一行）：

int main()
{
    bool a = 1;
    bool b = 1;
    bool c = 1;
    bool a_or_b = (a || b);
    bool not_a_or_b = !a_or_b;
    bool not_a_or_b__c = not_a_or_b || c;
    cout << "(a || b): " << (a || b) << '\n';
    cout << "!(a || b): " << !(a || b) << '\n';
    cout << "!(a || b) || c: " << (!(a || b)) || c << '\n';//HERE I'M GETTING 0 (incorrectly I would say)
    cout << "bool vars:\n";//WHY THIS LINE IS PRINTED AFTER THE PREVIOUS LINE BUT NOT BELOW IT?
    cout << "(a || b): " << a_or_b << '\n';
    cout << "!(a || b): " << not_a_or_b << '\n';
    cout << "!(a || b) || c: " << not_a_or_b__c << '\n';//HERE I'M GETTING 1
    return 0;
}

Answer 1

这是因为<<的优先级高于||。使用括号对其进行分组。

cout << "!(a || b) || c: " << ((!(a || b)) || c) << '\n';
//                            ^                ^

Answer 2

它解释

(!(a || b)) || c << '\n'

作为

(!(a || b)) || (c << '\n')

Answer 3

更改此部分：

<< (!(a || b)) || c << '\n'; //interpreted as (!(a || b)) || (c << '\n')

到此：

((!(a || b)) || c) << '\n'; //interpreted as intended!

Answer 4

这是一个钳位问题

你使用（！（a || b））|| c第一部分是0肯定也许解释者甚至不看[|] c部分。