一行代码中有多个逻辑运算符

问题描述

我正在Stack Overflow上搜索此问题的答案，但我没有找到确切的答案.我想出了这段代码.我知道操作员应该如何工作，但在这种问题上我不理解他们.例如，在第一种情况下，如果我正在使用++y和++z，z和y怎么仍为1?

I was searching on Stack Overflow for the answer to this question but I haven't found an exact answer. I came up with this code. I know how operators are supposed to work but I don't understand them in this kind of problem. For example, in the first case, how can z and y still be 1 if there I am using ++y and ++z?

#include <stdio.h>

int main(void) {
    int x, y, z;

    x = y = z = 1;
    ++x || ++y && ++z;
    printf("x = %d y = %d z = %d\n", x, y, z);

    x = y = z = 1;
    ++x && ++y || ++z;
    printf("x = %d y = %d z = %d\n", x, y, z);

    x = y = z = 1;
    ++x && ++y && ++z;
    printf("x = %d y = %d z = %d\n", x, y, z);

    x = y = z = -1;
    ++x && ++y || ++z;
    printf("x = %d y = %d z = %d\n", x, y, z);

    x = y = z = -1;
    ++x || ++y && ++z;
    printf("x = %d y = %d z = %d\n", x, y, z);

    x = y = z = -1;
    ++x && ++y && ++z;
    printf("x = %d y = %d z = %d\n", x, y, z);

    return 0;
}

结果我得到:

x = 2 y = 1 z = 1
x = 2 y = 2 z = 1
x = 2 y = 2 z = 2
x = 0 y = -1 z = 0
x = 0 y = 0 z = -1
x = 0 y = -1 z = -1

推荐答案

这是对逻辑表达式求值的结果:一旦确定表达式为假(或真)，其余运算符即为不再进行评估.例如:

This is a result of the evaluation of the logical expressions: as soon as it has been determined that an expression is false (or true), the remaining operators are not evaluated anymore. E.g.:

++x || ++y && ++z;

由于x为1，所以表达式将为true，而与z或y无关，因此不再执行++y和++z.

As x is one, the expression will be true independent of what z or y are, so ++y and ++z are not performed anymore.

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