功能类型的Golang空接口{}

Question

An object of any type can be assigned to an empty interface. For example, we have the following function

func Println(i interface{} ) {
  fmt.Println(i)
}

We can call it by

Println(3)
Println(1.5)
Println("Hello")

But I can't achieve the same thing for function type

func Map(fn func( interface{} )) {  
  ......  
}  

I cannot call this with
Map( func( i int) {......} )
because the type func(int) is different from the type func( interface{} ).

But when I define func( interface{} ), I really mean any type of the input parameters. How can I achieve this in Golang?

Answer 1

It fails because the signatures don't match.

When you call Println(3), the function isn't taking an integer as its first argument. Rather the integer gets packed inside an interface{} variable (an automatic conversion, since integers conform to the interface), and that variable is passed to the function. This conversion happens on the calling side, so the process of calling the function is different to calling a function matching func(i int).

If you want to write a function that accepts arbitrary unary functions, you will need to declare it to take an interface{} variable as its argument and then check the value using the reflect package. The reflect package can also help you call arbitrary functions where you don't know the signature at compile time.

For example:

func Map(f, v interface{}) interface{} {
    fn := reflect.ValueOf(f)
    fnType := fn.Type()
    if fnType.Kind() != reflect.Func || fnType.NumIn() != 1 || fnType.NumOut() != 1 {
        panic("Expected a unary function returning a single value")
    }
    res := fn.Call([]reflect.Value{reflect.ValueOf(v)})
    return res[0].Interface()
}

This will call the given function f with the argument v and return the result. Provided v is assignable to f's first argument the call will succeed without a panic. You can experiment with this example here: http://play.golang.org/p/kkBu56JYb8