函数参数的模式匹配 [英] Pattern matching on function parameters

问题描述

我有一个元组向量

val l = for {i <- 1 to 5} yield (i,i*2)
Vector((1,2), (2,4), (3,6), (4,8), (5,10))

我想用以下方式总结:

l.reduce((x,y) => (x._1+y._1, x._2+y._2))
(15,30)

但使用模式匹配.

如果函数只有一个参数，我知道怎么做，即:l.map({case(a,b)=>a+b})，但我不能t 让它使用两个参数工作.这就是我试图做的:

I know how to do it if the function gets only one parameter, ie: l.map({case(a,b)=>a+b}), but I can't get it to work with two parameters. this is what I tried to do:

l.reduce({(case(a,b),case(c,d))=>(a+c,b+d)})

但这行不通.

所以我的问题是，如何解包作为函数参数出现的 2 个元组?

So my question is, how can I unpack 2 tuples that come as a function parameters?

推荐答案

只需明确指定类型即可帮助偏函数机制:

Just specify type explicitly to help partial functions mechanism:

l.reduce[(Int, Int)]{ case ((a,b), (c, d)) => (a + b, c + d)}

没有 [(Int, Int)] Scala 无法为您的部分函数推断正确的类型

Without [(Int, Int)] scala can't infer correct type for your partial function

附言如果您想知道为什么会看到关于 String 的错误，而不管您实际使用的是什么类型(在您的情况下是 Int):

P.S. If you're intersesting why you see this error about String regardless what type you're actually using (it's Int in your case):

found   : Any , 
required: String

那是因为Predef 中的字符串隐含+".这不是这种隐式的唯一问题 - 您可以查看 SI-194.

That's because of implicit "+" for strings in Predef. It's not the only problem with this implicit - you may look at SI-194.

即使没有:

l.reduce{ case ((a,b), (c, d)) => (a, d)}

The argument types of an anonymous function must be fully known. (SLS 8.5)
Expected type was: (?, ?) => ?

scala 无法推断类型，因为它无法推断部分函数的类型 - 可能是 Int 的任何超类型:(A1, A1) =>A1 预期，其中 [A1 >: A]

scala was unable to infer the type, because it can't infer a type for partial function - may be any supertype of Int: (A1, A1) => A1 expected, where [A1 >: A]

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