时髦名单传播 [英] groovy list spreading

问题描述

如果运行以下Groovy代码，断言将通过

If you run the following Groovy code, the assertion passes

def foo(a, b) {
  a + b
}

assert 'aaabbb' == foo(['aaa', 'bbb'])

这表明如果使用包含X元素的List参数调用某个方法，那么List将被传播，并且将调用一个带有X参数的方法。

This suggests that if a method is called with a List parameter that contains X elements, then the List will be spread and a method with X arguments will be invoked.

当然，这只会在没有使用List（或其祖先）类型的单个参数定义的方法时才会发生。

Of course, this will only happen if there isn't a method defined with a single parameter of type List (or ancestor thereof).

在阅读另一篇SO Groovy答案时，我最近才发现这一点。我从未在Groovy文档，发行说明或书籍中看到过它。它是一个隐藏的功能，一个错误，或只是我错过了什么？

I only discovered this quite recently when reading another SO Groovy answer. I've never seen it mentioned in the Groovy docs, release notes, or books. Is it a hidden feature, a bug, or just something I've missed?

推荐答案

显然将在Groovy 2中被删除：
http：// groovy。 329449.n5.nabble.com/removing-features-in-Groovy-2-td4422494.html

Going to be removed in Groovy 2 apparently: http://groovy.329449.n5.nabble.com/removing-features-in-Groovy-2-td4422494.html

JT是第一个在删除名单上的似乎Groovy用户的每个人（充满感召力）都表示同意。

JT's first on the to-remove list and it seems everyone (with clout) on Groovy User agrees.

这篇关于时髦名单传播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！