A tale from The Go Programming Language Specification

August 23, 2021
go golang


The story started with this nice poll from @go101:

package main

import "reflect"

type T int

func (t T) M() { print(t) }

type S struct{ *T }

var t = new(T)
var s = S{T: t}

func main() {
	f := t.M
	g := s.M
	h := reflect.ValueOf(s).MethodByName("M").Interface().(func())
	*t = 5

The answer is 005, but it’s not the end of the story, since when @go101 stated that:

For the official Go compiler, the output is 005. “s.M” is promoted as “t.M” by the compiler. However, personally, I think the output is compiler depended. 000 and 055 should be also right answers. In fact, the two make the direct coding way and the reflection way more consistent.

My opinion is different, any Go implementation, includes compiler+runtime, must produce 005 to obey the Go specification.


Since then, I have a long dicussion with @go101, via some tweets and direct Twitter message. @go101’s conclusion is that:

s.M() is shorthand for s.T.M()


s.M is shorthand for s.T.M

We will find out why the Go specification mandate that behavior and the output of:

g := s.M

must be 0.

First, let look at the specification to see what is Method values:

If the expression x has static type T and M is in the method set of type T, x.M is called a method value. The method value x.M is a function value that is callable with the same arguments as a method call of x.M. The expression x is evaluated and saved during the evaluation of the method value; the saved copy is then used as the receiver in any calls, which may be executed later.

So in g := s.M, s.M is a selector expression, and from the spec:

For a value x of type T or *T where T is not a pointer or interface type, x.f denotes the field or method at the shallowest depth in T where there is such an f. If there is not exactly one f with shallowest depth, the selector expression is illegal.

Thus for s.M to be valid, we need to find where is M appears in S. And the answer is (*s.T).M. Further, the declared receiver parameter type of M is T, so s must be evaluated to something typed T during method value evaluation. And the answer is also *(s.T).

Here’s a simple program to prove that:

package main

import (

const input = `
package p

type T int

func (t T) M() { print(t) }

type S struct{ *T }

func (s S) N() {}

var t = new(T)
var s = S{T: t}

var fset = token.NewFileSet()

func main() {
	f, err := parser.ParseFile(fset, "main.go", input, 0)
	if err != nil {

	conf := types.Config{Importer: importer.Default()}
	pkg, err := conf.Check("p", fset, []*ast.File{f}, nil)
	if err != nil {

	styp := pkg.Scope().Lookup("S").Type()
	fmt.Printf("Method set of %s:\n", styp)
	ms := types.NewMethodSet(styp)
	fmt.Printf("%v\n", ms.Lookup(pkg, "M").Obj())
	fmt.Printf("%v\n", ms.Lookup(pkg, "N").Obj())

Run above program output:

Method set of p.S:
func (p.T).M()
func (p.S).N()

I think the only problem here is the lacking of example in the specification for method value receiver evaluation. There’re examples for method calls in the Selectors section:

p.M0()       // ((*p).T0).M0()      M0 expects *T0 receiver
p.M1()       // ((*p).T1).M1()      M1 expects T1 receiver

Though we are stay here, honestly, I still don’t understand why @go101 has this conclusion. The method value s.M and method call s.M() both must evaluate s.M the same way, it’s just that in method value case, the function value is never called.


I opened an issue and sent a CL to add an example for method value evaluation, which I hope will eliminate the confusion.


Another fun story with Go and its community for me. How about you?

Thanks for reading so far.

Till next time!

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