语句z = ++ x || ++ y& ++ z的C运算符优先级 [英] Operator precedence in C for the statement z=++x||++y&&++z

问题描述

我正在研究运算符的优先级，但无法理解x的值如何变为2，y和z的值如何变为1

I was studying operator precedence and I am not able to understand how the value of x became 2 and that of y and z is 1

x=y=z=1;

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

计算结果为

x=2 y=1 z=1

推荐答案

++具有比||更高的优先级，因此，分配的整个RHS都归结为x的增量，并对真值求值值(1).

++ has higher priority than ||, so the whole RHS of the assignment boils down to an increment of x and an evaluation to a truth value (1).

z = ++x         ||  ++y&&++z;
    truthy (1)     never executed

这是因为++x的计算结果为true，并且第二个分支未执行. ++x是2，在布尔上下文中，其值为true或1. z取值为1，为您提供观察到的最终状态.

This is because ++x evaluates to true and the second branch is not executed. ++x is 2 which, in a boolean context, evaluates to true or 1. z takes the value of 1, giving you the observed final state.

这篇关于语句z = ++ x || ++ y& ++ z的C运算符优先级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！