Struct in struct embedding promoted methods question

Question

package main

import "fmt"

type type1 struct { //T
}

func (t1 type1) type1Meth1() {
    fmt.Printf("==> func (t1 type1) type1Meth1():\n Type: %T\n Value: %+v\n\n", t1, t1)
}

func (t1 *type1) type1Meth2() {
    fmt.Printf("==> func (t1 *type1) type1Meth2():\n Type: %T\n Value: %p\n Contains: %+v\n\n", t1, t1, t1)
}

func (t1 type1) type1Meth3() {
    fmt.Printf("==> func (t1 type1) type1Meth3():\n Type: %T\n Value: %+v\n", t1, t1)
}

type type2 struct { //S
    type1
}

func (t2 *type2) type1Meth3() {
    fmt.Printf("==> func (t2 *type2) type1Meth3(): Type: %T\n Value: %+v\n\n", t2, t2)
}
func main() {
    t2 := type2{}
    t2.type1Meth1() // type2 contains method set of type1
    t2.type1Meth2() // not sure, why this works? type2 does not have method set of *type1 (A)
    t2.type1Meth3() // type2 contains method set of type1. intercepted by embedding type type2 and called with *type2 receiver
}

Gives me:

$ go run embed-struct-in-struct.go
==> func (t1 type1) type1Meth1():
 Type: main.type1
 Value: {}

==> func (t1 *type1) type1Meth2():
 Type: *main.type1
 Value: 0x116be80
 Contains: &{}

==> func (t2 *type2) type1Meth3(): Type: *main.type2
 Value: &{type1:{}}

go version
go version go1.17.2 darwin/amd64

Not sure why call in (A) works? Documentation says: promoted methods are included in the method set of the struct as follows:

Given a struct type S and a defined type T, promoted methods are included in the method set of the struct as follows:

If S contains an embedded field T, the method sets of S and *S both include promoted methods with receiver T. The method set of *S also includes promoted methods with receiver *T. If S contains an embedded field *T, the method sets of S and *S both include promoted methods with receiver T or *T.

score 3 · Accepted Answer · 2021-11-01 00:09:51Z

3

(A) works because the method call implicitly takes the address of the receiver.

The method (*type1).type1Meth2() is promoted to *type2 per the promotion rules quoted in the question.

The section of the specification on calls says:

A method call x.m() is valid if the method set of (the type of) x contains m and the argument list can be assigned to the parameter list of m. If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m():

The expression t2.type1Meth2() is shorthand for (&t2).type1Meth2() because t2 is addressable and type1Meth2() is in the method set of *type2.

edited Nov 1, 2021 at 0:09

answered Oct 31, 2021 at 22:03

user5728991

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1 Comment

Ankur Agarwal Over a year ago

wow interesting. The rules for method sets inclusion are not straightforward.

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