问题
Welcome to Scala version 2.10.2
Type in expressions to have them evaluated.
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scala> val fn = (x:Int) => x+1
fn: Int => Int = <function1>
scala> val fn1 = fn _
fn1: () => Int => Int = <function0>
scala> val fn2 = fn1 _
fn2: () => () => Int => Int = <function0>
I don't understand why the placeholder(without a suggested type) application to a function is creating a new curried function with prefixed additional void argument.
I was expecting a compiler error or at least a warning.
回答1:
I guess it's so due to the uniform access principle: in REPL val
is a field of object, not a local variable. And all not private[this]
fields are getter methods.
So your code is something like this:
def fn() = (x:Int) => x+1
val fn1 = fn _ // () => fn()
It works as expected with local variables:
scala> {
| val fn = (x:Int) => x+1
| val fn1 = fn _
| }
<console>:10: error: _ must follow method; cannot follow Int => Int
val fn1 = fn _
^
Even though I can explain why it works this way, I still think this behavior can be considered a error.
来源:https://stackoverflow.com/questions/17484457/unable-to-understand-placeholder-behavior-in-scala-function-application