短路和一元运算符的工作 [英] Working of short circuit and unary operator

问题描述

请查看以下代码:

int i=5;
boolean b = i<5 && ++i<5;//line 2
System.out.println(i);//line 3, prints 5

根据我的理解，在第2行中:由于在所有运算符中，++的优先级最高，因此应首先评估++i.但是line 3实际上正在打印i=5(而不是6).含义&&在++运算符之前进行了评估.这怎么可能?

In line 2, according to my understanding: Since among all the operators, ++ has highest precedence ++i should be evaluated first. But line 3 actually is printing i=5 (and not 6). Meaning, && has evaluated before ++ operator. How is it possible?

从答案中我看到在Java中，所有表达式都是从左到右求值的.". 但是实际上优先顺序何时生效.在以下代码中:

From the answers I see that "In Java, all expressions are evaluated from left to right.". But when does actually precedence order comes into play. In following code:

int a=1,b=1,c=1;
boolean b = a==b&&b==c;//Line2

在第2行中，代码不会只是从左到右运行.首先评估a == b，然后评估b == c，然后评估&&.操作员.你能解释更多吗?

In line2 code would't just run from left to right. First a==b is evaluated then b==c and then && operator. Can you please explain more?

推荐答案

那不是表达式的处理方式.

That's not how the expression is processed.

在Java中，所有表达式从左到右求值.仅当考虑对&&的自变量进行评估时，运算符优先级才起作用.

In Java, all expressions are evaluated from left to right. Operator precedence only comes into play when considering the evaluation of the arguments of &&.

所以i < 5是在甚至没有考虑++i < 5的情况下计算的.

So i < 5 is computed before ++i < 5 is even considered.

在这种情况下，将不评估++i < 5 ，因为i < 5是false.所以i停留在5.

In this case ++i < 5 will not be evaluated, since i < 5 is false. So i stays at 5.

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