C的短路评估 [英] Short-circuit evaluation on C

问题描述

我正在从Kelley-Pohl撰写的《关于C的书》中学习C，但是我不理解这种练习:

I'm studying C from A Book on C by Kelley-Pohl, and there's this exercise that I don't understand:

int a = 0, b = 0, x;

x = 0 && (a = b = 777);
printf("%d %d %d\n", a, b, x);
x = 777 || (a = ++b);
printf("%d %d %d\n", a, b, x);

他们只是说想想输出，然后将其与真实输出进行比较.我以为输出应该是

They just say to imagine the output and compare it to the real one. I thought the output would have been

777 777 0

777 777 0

778 778 1

778 778 1

但是是

0 0 0

0 0 0

0 0 1

推荐答案

&&运算符使用惰性评价.如果&&运算符的任一侧为false，则整个表达式为false.

The && operator uses lazy evaluation. If either side of the && operator is false, then the whole expression is false.

C检查运算符左侧的真实值，在您的情况下为0.由于0在c中为false，因此永远不会评估操作的右侧表达式(a = b = 777).

C checks the truth value of the left hand side of the operator, which in your case is 0. Since 0 is false in c, then the right hand side expression of the operation, (a = b = 777), is never evaluated.

第二种情况类似，不同之处在于，如果左侧表达式返回true，则||返回true.还要记住，在c中，任何不是0的东西都被认为是true.

The second case is similar, except that || returns true if the left hand side expression returns true. Also remember that in c, anything that is not 0 is considered true.

希望这会有所帮助.

这篇关于C的短路评估的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！