Question

I m seeing something like this in Scala code:

if (Set("foo", "bar").contains(x)) { ... }

一种明显的替代办法是,一种简单的OR条件:

if (x == "foo" || x == "bar") { ... }

除了可论证的更好的可读性外,使用原始方法的客观好处是什么? 是否有更多的表演者/消费较少的记忆?

注:

In the real code, "foo" and "bar" are fixed instances of a case class
There are just two elements in the condition. The likelihood of adding more elements in future is low.

Answer 1

我建议采用前一种做法的唯一原因是,如果这套方法具有某种可图性,或者如果它拥有如此庞大和完全的复杂数据,它实际上会有助益。

If that was the case, I would still give it a name and leverage the fact that contains is also aliased as apply for Sets to make it more readable (and avoid creating a lot of very short-lived data):

val isFooOrBar: String => Boolean = Set("foo", "bar")

if (isFooOrBar(x)) { ... }

关于业绩:

there is a highly likelihood that this is not critical
if it is critical, the only source of truth is measuring

在这方面,我确信,“很有可能取得更好的效果(没有数据可生成、低层抽象化、对时间很熟悉和JIT等),但即便如此,我还是将侧重于可读性。如果<条码>类别相当庞大,你可能不想将整件内容放在<条码>中,但不妨碍你做以下工作:


def isFooOrBar(x: String): Boolean = x == "foo" || x == "bar"

if (isFooOrBar(x)) { ... }

which makes it indistinguishable at call-site what approach you took and is arguably readable anyway. Maybe the only improvement is to give the predicate a name that described more accurately the why, like in the following made-up snippet:
// anything but "foo" and "bar" are incompatible with the retroencabulator
def isCompatibleWithRetroencabulator(s: String) = ...

如果你在上注意履约情况,就将其改为 私人,并可能将其附在@inline(及衡量标准);

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