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scala 下划线解析报错： missing parameter type for expanded function

本文首先介绍下eta-expansion概念，以及其对下划线的expand的解析规则和匿名函数简写注意事项，最后通过例子解析加深读者的理解

eta-expansion概念：

把 x => func(x) 简化为 func _ 或 func 的过程称为 eta-conversion?

把 func 或 func _ 展开为 x => func(x) 的过程为 eta-expansion

Eta?Expansion的就近expand解析原则：

Underscores extend outwards to the closest closing?Expr?: top-level expressions or expressions in parentheses

?

匿名函数简写注意事项：

所以当匿名函数有多个括号嵌套的时候，不要使用_的简写方式 ,而应该用普通匿名函数的书写方式。比如（x：Int，y:Int）=> （x*2）+f(y+2)

例子解析：

例子1：

scala>?List(1,2,3).map(_*2)

res6:?List[Int]?=?List(2,?4,?6)

scala>?List(1,2,3).map(_*2+1)

res14:?List[Int]?=?List(3,?5,?7)

但是吧_*2 换成(_*2) 则出错，因为根据就近原则?_*2 会在最近的括号中解析成匿名函数为(x$1)?=>?x$1.$times(2), ?变成为（x=>x*2) +1 , 而期望的是（x）=>(x*2)+1

scala>?List(1,2,3).map((_*2)+1)

<console>:8:?error:?missing?parameter?type?for?expanded?function?((x$1)?=>?x$1.$times(2))

??????????????List(1,2,3).map((_*2)+1)

???????????????????????????????^

如果括号内只有_ 本身，则_本身构不成一个表达式expressions，所以如下是允许的

scala>?List(1,2,3).map((_)+1)

res8:?List[Int]?=?List(2,?3,?4)

例子2：

scala>?List(1,2,3,4).foreach(print?_)

1234

_本身构不成一个表达式:expressions，所以如下是允许的

scala>?List(1,2,3,4).foreach(print(_))

1234

(_.toString）是一个在括号（）里的表达式，会expand解析成匿名函数，而print期望的String类型的值，所以编译报错

scala>?List(1,2,3,4).foreach(print(_.toString))

<console>:8:?error:?missing?parameter?type?for?expanded?function?((x$1)?=>?x$1.toString)

??????????????List(1,2,3,4).foreach(println(_.toString))

????????????????????????????????????????????^

例子3：

val myStrings = new Array[String](3)
// do some string initialization

// this works
myStrings.foreach(println(_))


// ERROR: missing parameter type for expanded function
myStrings.foreach(println(_.toString))

It expands to:

myStrings.foreach(println(x => x.toString))

You want:

myStrings.foreach(x => println(x.toString))

The placeholder syntax for anonymous functions replaces the smallest possible containing expression with a function.

keywords：error:?missing?parameter?type?for?expanded?function?

Eta?Expansion

The usage of?placeholder syntax for anonymous functions is restricted to expressions.

参考如下：

This has already been addressed in?a related question. Underscores extend outwards to the closest closing?Expr?: top-level expressions or expressions in parentheses.

(_.toString)?is an expression in parentheses. The argument you are passing to?Exception?in the error case is therefore, after expansion, the full anonymous function?(x$1) => x$1.toString?of type?A <: Any => String, while?Exception?expects a?String.

In the?println?case,?_?by itself isn‘t of syntactic category?Expr, but?(println (_))?is, so you get the expected?(x$0) => println(x$0).

http://stackoverflow.com/questions/4701761/why-do-i-get-a-missing-parameter-for-expanded-function-in-one-case-and-not-the

文章来自：http://zhouchaofei2010.iteye.com/blog/2260107