OR和AND在C温度范围 [英] OR and AND operation in C

问题描述

我在下面的程序一个疑问。

I have a doubt in the program below.

int main()
{
    int i = -3,j = 2, k = 0,m;
    m = ++i || ++j && ++k;
    printf("%d %d %d %d\n", i, j, k, m);
    return 0;
}

我得到的输出 -2 2 0 1 。

在或操作，如果第一个值是真实的，那么它不会评价：第二个让 I = -2 和 J = 2 。
然后是与运算。它会检查这两个值，如果被true.So K = 1 然后 M = 1 。
所以输出应该是 -2 2 1 1 。我运行和检查，并得到为输出-2 2 0 1 ，但我不明白为什么。

In OR operation if 1st value is true then it won't evaluate the 2nd one so i = -2 and j =2. Then comes the AND operation . It will check for both the value to be true.So if k = 1 then m = 1. So the output should be -2 2 1 1. I run and check and got output as -2 2 0 1 but I could not understand how.

推荐答案

您使用的短路或。因为++ i的计算结果为-2，这是不为0，短路和不评估前pression的其余部分。其结果是，既不J或K得到递增。

You used a short circuit or. Since ++i evaluates to -2, which is not 0, it short circuits and doesn't evaluate the rest of the expression. As a result, neither j or k get incremented.

还要注意的是短路的运营商，||和＆功放;＆放大器;，是左关联的||高precedence比＆安培;＆放;.其结果是，在||如果左侧的值为true，首先得到评估，及早出局，而与放大器;＆安培;早出局，如果左侧的值为false。

Also note that the short circuit operators, || and &&, are left associative and that || is higher precedence than &&. As a result, the || gets evaluated first, and early outs if the left hand side evaluates to true, while && early outs if the left hand side evaluates to false.

编辑：修正了解释precedence错误

Fixed a mistake with explaining the precedence.

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