Java有懒惰的评估吗？ [英] Does Java have lazy evaluation?

问题描述

我知道Java在这种情况下有智能/懒惰评估：

I know that Java has smart/lazy evaluation in this case:

public boolean isTrue() {
    boolean a = false;
    boolean b = true;
    return b || (a && b); // (a && b) is not evaluated since b is true
}

但是怎么样：

public boolean isTrue() {
    boolean a = isATrue();
    boolean b = isBTrue();
    return b || a;
}

isATrue（）调用即使 isBTrue（）返回true？

Is isATrue() called even if isBTrue() returns true?

推荐答案

在Java中（和其他类C语言），这被称为 短路评估 。 ^*

In Java (and other C-like languages), this is referred to as short-circuit evaluation.^*

是的，在第二个例子中 isATrue 总是被称为。也就是说，除非编译器/ JVM可以确定它没有可观察到的副作用，在这种情况下它可能会选择优化，但在这种情况下你无论如何都不会注意到差异。

And yes, in the second example isATrue is always called. That is, unless the compiler/JVM can determine that it has no observable side-effects, in which case it may choose to optimize, but in which case you wouldn't notice the difference anyway.

---

* 两者截然不同;前者本质上是一种优化技术，而第二种是由语言强制执行，可能影响可观察的程序行为。

我最初建议这是非常明显的来自懒惰的评价，但正如@Ingo在下面的评论中指出的那样，这是一个可疑的断言。人们可以将Java中的短路运算符视为惰性求值的一个非常有限的应用。

I originally suggested that this was quite distinct from lazy evaluation, but as @Ingo points out in comments below, that's a dubious assertion. One may view the short-circuit operators in Java as a very limited application of lazy evaluation.

然而，当函数式语言强制执行惰性求值语义时，它通常用于不同的原因，即防止无限（或至少，过度）递归。

However, when functional languages mandate lazy-evaluation semantics, it's usually for a quite different reason, namely prevention of infinite (or at least, excessive) recursion.

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