这是哪里的法律使用红宝石图示运营商？ [英] Where is it legal to use ruby splat operator?
问题描述
提示图标都很酷。他们不只是爆炸阵列，尽管这是乐趣。他们还可以转换为数组和扁平阵列（请参见 http://github.com/mischa/splat/tree/master为他们做什么的详尽清单。）
Splats are cool. They're not just for exploding arrays, although that is fun. They can also cast to Array and flatten arrays (See http://github.com/mischa/splat/tree/master for an exhaustive list of what they do.)
它看起来像一个人不能在图示执行其他操作，但在1.8.6 / 1.9以下code抛出意外tSTAR
It looks like one cannot perform additional operations on the splat, but in 1.8.6/1.9 the following code throws "unexpected tSTAR":
富=酒吧|| * ZAP＃=＆GT;意外tSTAR
而这个作品：
富= * ZAP ||酒吧＃=＆GT;工作，但价值有限
在哪里可以提示图标出现在一个前pression？
Where can the splat appear in an expression?
推荐答案
首先，precedence是不是一个问题在这里，因为富=酒吧|| （* ZAP）工作没有更好的。一般的经验法则是，你不能在图示执行其他操作。即使是一些如 = foo的简单（* ZAP）是无效的。这适用于1.9为好。
First, precedence isn't an issue here, because foo = bar || (*zap) works no better. The general rule of thumb is that you cannot perform additional operations on a splat. Even something as simple as foo = (*zap) is invalid. This applies to 1.9 as well.
说了这么多，你能指望什么富=酒吧|| * ZAP 做的，如果它的工作，比富=酒吧不同|| ZAP ？即使是在像 A，B =酒吧案例|| * ZAP （这也不起作用）， A，B = ||吧ZAP 完成什么，我会认为是一样的。
Having said that, what do you expect foo = bar || *zap to do, if it worked, that is different than foo = bar || zap? Even in a case like a, b = bar || *zap (which also doesn't work), a, b = bar || zap accomplishes what I'd assume would be the same thing.
在哪里，这可能使任何意义，唯一的情况是一样的东西 A，B = foo的，酒吧|| * ZAP 。你会发现，在这里，你会想用这个大多数情况下是由覆盖A，B = foo，那么*（巴|| ZAP）。如果不包括你的情况，你应该问问自己，你真的希望通过写这样一个丑陋的结构来完成。
The only situation where this might make any sense is something like a, b = foo, bar || *zap. You should find that most cases where you would want to use this are covered by a, b = foo, *(bar || zap). If that doesn't cover your case, you should probably ask yourself what you really hope to accomplish by writing such an ugly construct.
在回答您的意见， * ZAP ||酒吧等同于 *（ZAP ||吧）。这表明图示的precedence有多低。究竟有多低呢？最好的答案，我可以给你的是pretty低。
In response to your comments, *zap || bar is equivalent to *(zap || bar). This demonstrates how low the splat's precedence is. Exactly how low is it? The best answer I can give you is "pretty low".
有关一个有趣的例子，不过，考虑的方法富这三个参数：
For an interesting example, though, consider a method foo which takes three arguments:
def foo(a, b, c)
#important stuff happens here!
end
美孚（*巴= [1，2，3]）在转让之后将图示和参数分别设置为1，2和3。相比之下，与美孚（（*巴= [1，2，3]））这会抱怨具有错误的参数数目（1 3）。
foo(*bar = [1, 2, 3]) will splat after the assignment and set the arguments to 1, 2, and 3 respectively. Compare that with foo((*bar = [1, 2, 3])) which will complain about having the wrong number of arguments (1 for 3).
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