Scala 2.8：_行为改变了？

发布于 2024-08-30 16:37:16 字数 1577 浏览 2 评论 0原文

使用 XScalaWT，在 Scala 2.7 下编译：

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

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

使用 2.8.0 RC1，我收到此错误：

类型不匹配；发现：main.scala.NodeView 需要：org.eclipse.swt.widgets.Label

类型为：

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

所以看起来 _ 现在绑定到外部函数而不是内部函数。

这个改变是有意为之吗？

更新：这是一个最小化的示例：

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

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，这有效：

scala> foo(conv(println _))
1

原文

Using XScalaWT, this compiled under Scala 2.7:

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

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

With 2.8.0 RC1, I get this error:

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

The types are:

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

So it looks like _ now binds to the outer function instead of inner.

Is this change intentional?

UPDATE: Here is a minimized example:

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

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

Interestingly enough, this works:

scala> foo(conv(println _))
1

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眼眸 2024-09-06 16:37:16

以下是我从 Lukas Rytz 在 scala 用户列表中得到的答案：

嗨阿列克谢，
语义发生了变化
当我们引入命名参数时。一个
表达方式
<前><代码>foo(a = _)
过去的解析方式如下：
foo(x => a = x)
在 2.8 中，“a”被视为命名的
参数，即表达式是
解析为：
<前><代码>x =>富（a = x）
我将添加一个迁移警告
这一变化。

Here is the answer I got from Lukas Rytz on the scala-user list:

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.