Scala方法,它将函数作为参数然后执行它

Question

我想编写一个简单的方法,将函数作为参数然后执行它.

def exec(f: (a:Int, b:Int) => Boolean): Boolean = f(a,b)

我不确定上面有什么问题,但我收到错误:

<console>:1: error: ')' expected but ':' found.

Answer 1

def exec(f: (a:Int, b:Int) => Boolean): Boolean = f(a,b)
              ^      ^
              |      |
//   These are supposed to be types, but a: Int and b: Int aren't types,
//   they are identifiers with type ascriptions.

它看起来应该更像:

def exec(f: (Int, Int) => Boolean): Boolean = f(a, b)

现在f是一个功能(Int, Int) => Boolean.但这不会编译,因为a并b没有定义.

您需要传入它们,或将它们修复为值.

def exec(a: Int, b: Int)(f: (Int, Int) => Boolean): Boolean = f(a, b)

scala> exec(2, 3)(_ > _)
res1: Boolean = false