在Scala中是否可以有一个将方法作为参数的通用方法？

Question

I would like to have:

def myMethod[T < ???](f: T): Unit = {
    f()
}

The rest of the method is not really important but is it possible to replace ??? by somethind which would make sure T is a method

and if possible to go even further and make sure the return type of T is something defined ?

Like [T < (_*) => Int]

Thank you.

Answer 1

Would defining a type , like in the following trivial example, address your need?

type Fcn = String => Int

def myMethod(s: String, f: Fcn): Unit = {
  println(f(s))
}

myMethod("hello", (s: String) => s.length)
// 5

Answer 2

You can use function literals:

def myMethod[A, T](f: A => T) {
  f(someAValue)
}

or if you want functions that take no arguments:

def myMethod[T](f: () => T) {
  f()
}

But judging by your comments, it seems like you specifically want to reference methods, and query the method for information about itself, which is not a feature of basic Scala. You may be able to do some of that with reflection, but reflection is an area best avoided.

Answer 3

Technically, you can't directly pass a method, since a method is not a value that you can instantiate and pass-around. What is usually passed around is a function instance. And when it appears that you're passing a method, the compiler is actually creating a function instance for you (method lifting or eta-expansion).

But that is not going to work if you're looking to inspect meta data about that method (name, deprecation). For that, you're probably looking for reflection. You can have your method take an instance of java.reflect.Method but that will require that you obtain that at call-site.