Golang a map of functions with a receiver

Question

Is there anyway to make a map of function pointers, but functions that take recievers? I know how to do it with regular functions:

package main

func someFunc(x int) int {
    return x
}

func main() {
    m := make(map[string]func(int)int, 0)
    m["1"] = someFunc
    print(m["1"](56))
}

But can you do that with functions that take recievers? Something like this (though I've tried this and it doesn't work):

package main

type someStruct struct {
    x int
}

func (s someStruct) someFunc() int {
    return s.x
}

func main() {
    m := make(map[string](someStruct)func()int, 0)
    s := someStruct{56}
    m["1"] = someFunc
    print(s.m["1"]())
}

An obvious work around is to just pass the struct as a parameter, but that's a little dirtier than I would have liked

No - the methods aren't free-floating functions, you can't have a reference to a method without an associated instance, and you can't then execute it against a different instance. You could do this using reflection. — Adrian
– Adrian, Commented Aug 3, 2017 at 0:30
You actually can have a reference for a method without an associated instance: golang.org/ref/spec#Method_expressions. See my answer for an example. — eugenioy
– eugenioy, Commented Aug 3, 2017 at 2:17

eugenioy · Accepted Answer · 2017-08-03 02:16:14Z

7

You can do that using Method Expressions:

https://golang.org/ref/spec#Method_expressions

The call is a bit different, since the method expression takes the receiver as the first argument.

Here's your example modified:

package main

type someStruct struct {
    x int
}

func (s someStruct) someFunc() int {
    return s.x
}

func main() {
    m := make(map[string]func(someStruct)int, 0)
    s := someStruct{56}
    m["1"] = (someStruct).someFunc
    print(m["1"](s))
}

And here's a Go playground for you to test it:

https://play.golang.org/p/PLi5A9of-U

answered Aug 3, 2017 at 2:16

eugenioy

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