斯卡拉2.8：行爲改變了嗎？

Question

使用XScalaWT，這斯卡拉2.7下編譯：

class NodeView(parent: Composite) extends Composite(parent) { 
    var nodeName: Label = null 

    this.contains(
    label(
     nodeName = _ 
    ) 
) 
} 

隨着2.8.0 RC1，我得到這個錯誤：

type mismatch; found : main.scala.NodeView required: org.eclipse.swt.widgets.Label

的類型是：

label(setups: (Label => Unit)*)(parent: Composite) : Label 
contains(setups: (W => Unit)*) : W 

所以看起來像_現在綁定到外部函數而不是內部。

這種改變是故意的嗎？

UPDATE：這是一個最小化的例子：

的Scala 2.7.7：

scala> var i = 0 
i: Int = 0 

scala> def conv(f: Int => Unit) = if (_:Boolean) f(1) else f(0)  
conv: ((Int) => Unit)(Boolean) => Unit 

scala> def foo(g: Boolean => Unit) { g(true) }  
foo: ((Boolean) => Unit)Unit 

scala> foo(conv(i = _))  

scala> i  
res4: Int = 1 

的Scala 2.8.0RC3：

scala> var i = 0 
i: Int = 0 

scala> def conv(f: Int => Unit) = if (_:Boolean) f(1) else f(0) 
conv: (f: (Int) => Unit)(Boolean) => Unit 

scala> def foo(g: Boolean => Unit) { g(true) } 
foo: (g: (Boolean) => Unit)Unit 

scala> foo(conv(i = _)) 
<console>:9: error: type mismatch; 
found : Boolean 
required: Int 
     foo(conv(i = _)) 
       ^

scala> foo(conv(j => i = j)) 

scala> i 
res3: Int = 1 

有趣的是，這個作品：

scala> foo(conv(println _)) 
1 

Answer 1

下面是我從盧卡斯Rytz得到了斯卡拉用戶列表上的答案：

Hi Alexey,

there has been a change in semantics when we introduced named arguments. An expression

foo(a = _) 

used to be parsed as follows:

foo(x => a = x) 

In 2.8, "a" is treated as a named argument, i.e. the expression is parsed as:

x => foo(a = x) 

I will add a migration warning for this change.

Answer 2

我發現同樣的事情，並要求在Dave的博客：

http://www.coconut-palm-software.com/the_new_visual_editor/doku.php?id=blog:simplifying_swt_with_scala#comment__930ba2f0a020203873d33decce01ebc2

沒有回答有尚未雖然。就像你所說的那樣，它似乎像綁定到外部封閉而不是內部封閉。

更改

nodeName = _ 

到

x => nodeName = x 

解決了這個問題。