Type Embedding

From the article structs in Go, we know that a struct type can have many fields. Each field is composed of one field name and one field type. In fact, sometimes, a struct field can be composed of one field type only. The way to declare struct fields is called type embedding.

This article will explain the purpose of type embedding and all kinds of details in type embedding.

What Does Type Embedding Look Like?

Here is an example demonstrating type embedding:
package main

import "net/http"

func main() {
	type P = *bool
	type M = map[int]int
	var x struct {
		string // a defined non-pointer type
		error  // a defined interface type
		*int   // a non-defined pointer type
		P      // an alias of a non-defined pointer type
		M      // an alias of a non-defined type

		http.Header // a defined map type
	}
	x.string = "Go"
	x.error = nil
	x.int = new(int)
	x.P = new(bool)
	x.M = make(M)
	x.Header = http.Header{}
}

In the above example, six types are embedded in the struct type. Each type embedding forms an embedded field.

Embedded fields are also called as anonymous fields. However, each embedded field has a name specified implicitly. The unqualified type name of an embedded field acts as the name of the field. For example, the names of the six embedded fields in the above examples are string, error, int, P, M, and Header, respectively.

Which Types Can be Embedded?

The current Go specification (version 1.12) says
An embedded field must be specified as a type name T or as a pointer to a non-interface type name *T, and T itself may not be a pointer type.

The above description is accurate before Go 1.09. However, with the introduction of type aliases in Go 1.09, the description becomes a little outdated and inaccurate. For example, the description doesn't include the case of the P field in the example in the last section.

Here, the article tries to provide more accurate descriptions. Or, a slight modification on the description in Go specification:
An embedded field must be specified as the form T or *T, where T is a type name and the embedded type may not be a pointer type whose base type is a pointer or interface type.

The following lists some example types which can and can't be embedded:
type Encoder interface {Encode([]byte) []byte}
type Person struct {name string; age int}
type Alias = struct {name string; age int}
type AliasPtr = *struct {name string; age int}
type IntPtr *int
type AliasPP = *IntPtr

// These types and aliases can be embedded.
Encoder
Person
*Person
Alias
*Alias
AliasPtr
int
*int

// These types and aliases can't be embedded.
AliasPP          // base type is a pointer type
*Encoder         // base type is an interface type
*AliasPtr        // base type is a pointer type
IntPtr           // defined pointer type
*IntPtr          // base type is a pointer type
*chan int        // base type is a non-defined type
struct {age int} // non-defined non-pointer type
map[string]int   // non-defined non-pointer type
[]int64          // non-defined non-pointer type
func()           // non-defined non-pointer type

No two fields are allowed to have the same name in a struct, there are no exceptions for anonymous struct fields. By the embedded field naming rules, a non-defined pointer type can't be embedded along with its base type in the same struct type. For example, int and *int can't be embedded in the same struct type.

A struct type can't embed itself or its aliases, recursively.

Generally, it is only meaningful to embed types who have fields or methods (the following sections will explain why), though some types without any field and method can also be embedded.

What Is the Meaningfulness of Type Embedding?

The main purpose of type embedding is to extend the functionalities of the embedded types into the embedding type, so that we don't need to re-implement the functionalities of the embedded types for the embedding type.

Many other object-oriented programming languages use inheritance to achieve the same goal of type embedding. Both mechanisms have their own benefits and drawbacks. Here, this article will not discuss which one is better. We should just know Go chose the type embedding mechanism, and there is a big difference between the two: Here is an example to show how an embedding type extends the functionalities of the embedded type.
package main

import "fmt"

type Person struct {
	Name string
	Age  int
}
func (p Person) PrintName() {
	fmt.Println("Name:", p.Name)
}
func (p *Person) SetAge(age int) {
	p.Age = age
}

type Singer struct {
	Person // extends Person by embedding it
	works  []string
}

func main() {
	var gaga = Singer{Person: Person{"Gaga", 30}}
	gaga.PrintName() // Name: Gaga
	gaga.Name = "Lady Gaga"
	(&gaga).SetAge(31)
	(&gaga).PrintName()   // Name: Lady Gaga
	fmt.Println(gaga.Age) // 31
}

From the above example, it looks that, after embedding type Person, the type Singer obtains all methods and fields of type Person, and type *Singer obtains all methods of type *Person. Are the conclusions right? The following sections will answer this question.

Please note that, a Singer value is not a Person value, the following code doesn't compile:
var gaga = Singer{}
var _ Person = gaga

Does the Embedding Type Obtain the Fields and Methods of the Embedded Types?

Let's list all the fields and methods of type Singer and the methods of type *Singer used in the last example by using the reflection functionalities provided in the reflect standard package.
package main

import (
	"fmt"
	"reflect"
)

... // the types declared in the last example

func main() {
	t := reflect.TypeOf(Singer{}) // the Singer type
	fmt.Println(t, "has", t.NumField(), "fields:")
	for i := 0; i < t.NumField(); i++ {
		fmt.Print(" field#", i, ": ", t.Field(i).Name, "\n")
	}
	fmt.Println(t, "has", t.NumMethod(), "methods:")
	for i := 0; i < t.NumMethod(); i++ {
		fmt.Print(" method#", i, ": ", t.Method(i).Name, "\n")
	}

	pt := reflect.TypeOf(&Singer{}) // the *Singer type
	fmt.Println(pt, "has", pt.NumMethod(), "methods:")
	for i := 0; i < pt.NumMethod(); i++ {
		fmt.Print(" method#", i, ": ", pt.Method(i).Name, "\n")
	}
}
The result:
main.Singer has 2 fields:
 field#0: Person
 field#1: works
main.Singer has 1 methods:
 method#0: PrintName
*main.Singer has 2 methods:
 method#0: PrintName
 method#1: SetAge

From the result, we know that the type Singer really owns a PrintName method, and the type *Singer really owns two methods, PrintName and SetAge. But the type Singerdoesn't own a Name field. Then why is the selector expression gaga.Name legal for a Singer value gaga? Please read the next section to get the reason.

Shorthands of Selectors

From the articles structs in Go and methods in Go, we have learned that, for a value x, x.y is called a selector, where y is either a field name or a method name. If y is a field name, then x must be a struct value or a struct pointer value. A selector is an expression, which represents a value. If the selector x.y denotes a field, it may also has its own fields (if x.y is a struct value) and methods. Such as x.y.z, where z can also be either a field name or a method name.

In Go, (without considering selector colliding and shadowing explained in a later section), if a middle name in a selector corresponds to an embedded field, then that name can be omitted from the selector. This is why embedded fields are also called anonymous fields.

For example:
package main

type A struct {
	x int
}
func (a A) MethodA() {}

type B struct {
	A
}
type C struct {
	B
}

func main() {
	var c C

	// The following 4 lines are equivalent.
	_ = c.B.A.x
	_ = c.B.x
	_ = c.A.x
	_ = c.x // x is called a promoted field of type C

	// The following 4 lines are equivalent.
	c.B.A.MethodA()
	c.B.MethodA()
	c.A.MethodA()
	c.MethodA()
}

This is why the expression gaga.Name is legal in the example in the last section. For it is just the shorthand of gaga.Person.Name. Name is called a promoted field of type Singer.

As any embedding type must be a struct type, and the article structs in Go has mentioned that the field of an addressable struct value can be selected through the pointers of the struct value, so the following code is also legal in Go.
func main() {
	var c C
	pc = &c

	// The following 4 lines are equivalent.
	fmt.Println(pc.B.A.x)
	fmt.Println(pc.B.x)
	fmt.Println(pc.A.x)
	fmt.Println(pc.x)

	// The following 4 lines are equivalent.
	pc.B.A.MethodA()
	pc.B.MethodA()
	pc.A.MethodA()
	pc.MethodA()
}

Similarly, the selector gaga.PrintName can be viewed as a shorthand of gaga.Person.PrintName. But, it is also okay if we think it is not a shorthand. After all, the type Singer really has a PrintName method, though the method is declared implicitly (please read the section after next for details). For the similar reason, the selector (&gaga).PrintName and (&gaga).SetAge can also be viewed as, or not as, shorthands of (&gaga.Person).PrintName and (&gaga.Person).SetAge.

Note, we can also use the selector gaga.SetAge, only if gaga is an addressable value of type Singer. It is just syntactical sugar of (&gaga).SetAge. Please read method calls for details.

In the above examples, c.B.A.x is called the full form of selectors c.x, c.B.x and c.A.x. Similarly, c.B.A.MethodA is called the full form of selectors c.MethodA, c.B.MethodA and c.A.MethodA.

If every middle name in the full form of a selector corresponds to an embedded field, then the number of middle names in the selector is called the depth of the selector. For example, the depth of the selector c.MethodA used in an above example is 2, for the full form of the selector is c.B.A.MethodA.

Selector Shadowing and Colliding

For a value x, it is possible that many of its full-form selectors have the same last item y and every middle name of these selectors represents an embedded field. For such cases,

If a method selector is shadowed by another method selector, and the two corresponding method signatures are identical, we say the first method is overridden by the other one.

For example, assume A, B and C are three defined types.
type A struct {
	x string
}
func (A) y(int) bool {
	return false
}

type B struct {
	y bool
}
func (B) x(string) {}

type C struct {
	B
}
The following code doesn't compile. The reasons is selector v1.A.x collides with v1.B.x collide with each other, so both of them can't be shortened to v1.x. The same situation is for selector v1.A.y and v1.B.y.
var v1 struct {
	A
	B
}

func f1() {
	_ = v1.x
	_ = v1.y
}

The following code compiles okay. The selector v2.C.B.x is shadowed by v2.A.x, so the selector v2.x is a shortened form of v2.A.x actually. For the same reason, the selector v2.y is a shortened form of v2.A.y, not of v2.C.B.y.
var v2 struct {
	A
	C
}

func f2() {
	fmt.Printf("%T \n", v2.x) // string
	fmt.Printf("%T \n", v2.y) // func(int) bool
}

Implicit Methods for Embedding Types

As mentioned above, both of type Singer and type *Singer have a PrintName method each, and the type *Singer also has a SetAge method. However, we never explicitly declare these methods for the two types. Where do these methods come from?

In fact, assume a struct type S embeds a type T and the embedding is legal,

The above facts still hold true even if *T is not embeddable (a.k.a, T is a pointer or interface type), in which cases, the method set of *T is blank.

Simply speaking, if no selector shadowing and collisions happen, The following methods are implicitly declared by compilers for type Singer and type *Singer.
func (s Singer) PrintName() {
	s.Person.PrintName()
}

func (s *Singer) PrintName() {
	(*s).Person.PrintName()
}

func (s *Singer) SetAge(age int) {
	// <=> (&((*s).Person)).SetAge(age)
	(&s.Person).SetAge(age)
}

From the article methods in Go, we know that we can't explicitly declare methods for non-defined struct types and non-defined pointer types whose base types are non-defined struct types. But through type embedding, such non-defined types can also own methods.

Here is another example to show which implicit methods are declared.
package main

import "fmt"
import "reflect"

type F func(int) bool
func (f F) Validate(n int) bool {
	return f(n)
}
func (f *F) Modify(f2 F) {
	*f = f2
}

type B bool
func (b B) IsTrue() bool {
	return bool(b)
}
func (pb *B) Invert() {
	*pb = !*pb
}

type I interface {
	Load()
	Save()
}

func PrintTypeMethods(t reflect.Type) {
	fmt.Println(t, "has", t.NumMethod(), "methods:")
	for i := 0; i < t.NumMethod(); i++ {
		fmt.Print(" method#", i, ": ",
				t.Method(i).Name, "\n")
	}
}

func main() {
	var s struct {
		F
		*B
		I
	}

	PrintTypeMethods(reflect.TypeOf(s))
	fmt.Println()
	PrintTypeMethods(reflect.TypeOf(&s))
}
The result:
struct { main.F; *main.B; main.I } has 5 methods:
 method#0: Invert
 method#1: IsTrue
 method#2: Load
 method#3: Save
 method#4: Validate

*struct { main.F; *main.B; main.I } has 6 methods:
 method#0: Invert
 method#1: IsTrue
 method#2: Load
 method#3: Modify
 method#4: Save
 method#5: Validate

If a struct type embeds a type which implements an interface type (the embedded type may be the interface type itself), then generally the struct type also implements the interface type, exception there is a method specified by the interface type shadowed by or colliding other methods or fields. For example, in the above example program, both the embedding struct type and the pointer type whose base type is the embedding struct type implement the interface type I.

Please note, a type will only obtain the methods of the types it embeds directly or indirectly. For example, in the following code,
type MyInt int
func (mi MyInt) IsOdd() bool {
	return mi%2 == 1
}

type Age MyInt

type X struct {
	MyInt
}
func (x X) Double() MyInt {
	return x.MyInt + x.MyInt
}

type Y struct {
	Age
}

type Z X

Interface Types Embed Interface Types

Not only can struct types embed other types, but also can interface types. But interface types can only embed interface types. Please read interfaces in Go for details.

An Interesting Type Embedding Example

In the end, let's view an interesting example. The example program will dead loop and stack overflow. If you have understood the above content and polymorphism and type embedding, it is easy to understand why it will dead loop.
package main

type I interface {
	m()
}

type T struct {
	I
}

func main() {
	var t T
	var i = &t
	t.I = i
	i.m() // will call t.m(), then call i.m() again, ...
}


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