什么 - > < - 操作员吗？ [英] What does the -> <- operator do?

问题描述

我最近发现了以下代码：

I recently came upon the following code:

IntPredicate neg = x -> x <- x;

这是什么，某种反向双lambda？

What is this, some sort of reverse double lambda?

推荐答案

没有 - > < - 运算符。第一个 - > 只是lambda语法，如Java 8中所介绍的那样，第二个< - 是一个'小于'< 和'一元减去' - 的误导性连接。

There is no -> <- operator. That first -> is just lambda syntax, as introduced in Java 8, and that second <- is a misleading concatenation of 'smaller than' < and 'unary minus' -.

您可以将其读作 IntPredicate neg =（x） - > （x <（ - x））; ，即它测试 x 是否小于 -x ，这是所有人的情况（好吧，大多数）负数，因此名称 neg 。

You can read it as IntPredicate neg = (x) -> (x < (-x));, i.e. it tests whether x is smaller than -x, which is the case for all (well, most) negative numbers, hence the name neg.

IntPredicate neg = x -> x <- x;
System.out.println(neg.test(4));   // false
System.out.println(neg.test(0));   // false
System.out.println(neg.test(-4));  // true

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仅为完整性：此测试是不仅（故意？）难以理解，但是 - 正如评论中所指出的那样 - 它也失败了 Integer.MIN_VALUE （这是 == - Integer.MIN_VALUE的）。相反，你应该只使用更简单的 IntPredicate neg = x - > （x <0）; 。

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Just for completeness: This test is not only (intentionally?) hard to understand, but -- as pointed out in the comments -- it also fails for Integer.MIN_VALUE (which is ==-Integer.MIN_VALUE). Instead, you should probably just use the much simpler IntPredicate neg = x -> (x < 0);.

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