调用函数，无论其参数类型如何

Question

Suppose I have a function which resides in the fn property of the following Method struct:

type Method struct {
  fn interface{}
}

var inst = &Method{func(a, b int) int {
  return a + b
}}

Now, I want to invoke this function with two arguments without explicitly casting it to func(int, int) int like so

a := 5
b := 6
fmt.Println(inst.fn(a, b))

How can I achieve this? Is there some generic solution for this?

Answer 1

The only way I know is by using reflect.Value.Call:

type Method struct {
    fn interface{}
}

func (m Method) Call(args ...interface{}) {
    vs := make([]reflect.Value, len(args))
    for i := range args {
        vs[i] = reflect.ValueOf(args[i])
    }
    v := reflect.ValueOf(m.fn)
    v.Call(vs)
}

func main() {
    f := func(a, b int) {
        fmt.Println(a + b)
    }
    m := Method{f}
    m.Call(2, 3)
}

Playground: http://play.golang.org/p/JNtj2EMpu7.

Note: this will panic if fn is not a function or if the number or types of arguments are wrong. If you don't want that, you need to recheck all those conditions yourself.