Go supports some object-orient programming features. Method is one of these features. This article will introduce method related concepts in Go.
T and *T,
where T must satisfy 4 conditions:
T must be a defined type;
T must be defined in the same package as the method declaration;
T must not be a pointer type;
T must not be an interface type.
Interface types will be explained in
the next article.
Type T and *T are called the receiver type of the respective methods declared for them.
Type T is called the receiver base types of all methods declared for both type T and *T.
Note, we can also declare methods for type aliases
of the T and *T types specified above.
The effect is the same as declaring methods for the T
and *T types themselves.
If a method is declared for a type, we can say the type has (or owns) the method.
From the above listed conditions, we will get the conclusions that we can never (explicitly) declare methods for:int and string,
for we can't declare methods in the builtin standard package.
*T which are described above.
A method declaration is similar to a function declaration,
but it has an extra parameter declaration part.
The extra parameter part can contain one and only one parameter of the receiver type of the method.
The only one parameter is called a receiver parameter of the method declaration.
The receiver parameter must be enclosed in a ()
and declared between the func keyword and the method name.
// Age and int are two distinct types. We
// can't declare methods for int and *int,
// but can for Age and *Age.
type Age int
func (age Age) LargerThan(a Age) bool {
return age > a
}
func (age *Age) Increase() {
*age++
}
// Receiver of custom defined function type.
type FilterFunc func(in int) bool
func (ff FilterFunc) Filte(in int) bool {
return ff(in)
}
// Receiver of custom defined map type.
type StringSet map[string]struct{}
func (ss StringSet) Has(key string) bool {
_, present := ss[key]
return present
}
func (ss StringSet) Add(key string) {
ss[key] = struct{}{}
}
func (ss StringSet) Remove(key string) {
delete(ss, key)
}
// Receiver of custom defined struct type.
type Book struct {
pages int
}
func (b Book) Pages() int {
return b.pages
}
func (b *Book) SetPages(pages int) {
b.pages = pages
}
From the above examples, we know that the receiver base types not only can be struct types, but also can be other kinds of types, such as basic types and container types, as long as the receiver base types satisfy the 4 conditions listed above.
In some other programming languages,
the receiver parameter names are always the implicit this,
which is not a recommended identifier for receiver parameter names in Go.
The receiver of type *T is called pointer receiver,
non-pointer receivers are called value receivers.
Personally, I don't recommend to view the terminology pointer
as an opposite of the terminology value,
because pointer values are just special values.
But, I am not against using the pointer receiver and value receiver terminologies here.
The reason will be explained below.
Method names can be the blank identifier _.
A type can have multiple methods with the blank identifier as name.
But such methods can never be called.
Only exported methods can be called from other packages.
Method calls will be introduced in a later section.
Book and type *Book
in the last example in the last section,
two following functions are implicitly declared by compiler:
func Book.Pages(b Book) int {
// The body is the same as the Pages method.
return b.pages
}
func (*Book).SetPages(b *Book, pages int) {
// The body is the same as the SetPages method.
b.pages = pages
}
In each of the two implicit function declarations, the receiver parameter is removed from its corresponding method declaration and inserted into the normal parameter list as the first one. The function bodies of the two implicitly declared functions is the same as their corresponding method explicit bodies.
The implicit function names,Book.Pages and (*Book).SetPages,
are both of the form TypeDenotation.MethodName.
As identifiers in Go can't contain the period special characters,
the two implicit function names are not legal identifiers,
so the two functions can't be declared explicitly.
They can only be declared by compilers implicitly,
but they can be called in user code:
package main
import "fmt"
type Book struct {
pages int
}
func (b Book) Pages() int {
return b.pages
}
func (b *Book) SetPages(pages int) {
b.pages = pages
}
func main() {
var book Book
// Call the two implicit declared functions.
(*Book).SetPages(&book, 123)
fmt.Println(Book.Pages(book)) // 123
}
In fact, compilers not only declare the two implicit functions, they also rewrite the two corresponding explicit declared methods to let the two methods call the two implicit functions in the method bodies (at least, we can think this happens), just like the following code shows:
func (b Book) Pages() int {
return Book.Pages(b)
}
func (b *Book) SetPages(pages int) {
(*Book).SetPages(b, pages)
}
T,
a corresponding method with the same name will be
implicitly declared by compiler for type *T.
By the example above, the Pages method is declared for type Book,
so a method with the same name Pages is implicitly declared
for type *Book:
// Note: this is not a legal Go syntax.
// It is shown here just for explanation purpose.
// It indicates that the expression (&aBook).Pages
// is evaluated as aBook.Pages (see below sections).
func (b *Book) Pages = (*b).Pages
This is why I don't reject the use of the value receiver terminology (as the opposite of the pointer receiver terminology). After all, when we explicitly declare a method for a non-pointer type, in fact two methods are declared, the explicit one is for the non-pointer type and the implicit one is for the corresponding pointer type.
As mentioned at the last section, for each declared method, compilers will also declare a corresponding implicit function for it. So for the implicitly declared method, the following implicit function is declared by compiler.func (*Book).Pages(b *Book) int {
return Book.Pages(*b)
}
In other words, for each explicitly declared method with a value receiver, two implicit functions and one implicit method will also be declared at the same time.
A method specification can be viewed as a
function prototype
without the func keyword.
We can view each method declaration is composed of the func keyword,
a receiver parameter declaration, a method specification and a method (function) body.
Pages and
SetPages methods shown above are
Pages() int
SetPages(pages int)
Each type has a method set.
The method set of a non-interface type is composed of all the method specifications
of the methods declared, either explicitly or implicitly, for the type,
except the ones whose names are the blank identifier _.
Interface types will be explained in
the next article.
Book type shown in the
previous sections is
Pages() int
*Book type is
Pages() int
SetPages(pages int)
The order of the method specifications in a method set is not important for the method set.
For a method set, if every method specification in it is also in another method set, then we say the former method set is a subset of the latter one, and the latter one is a superset of the former one. If two method sets are subsets (or supersets) of each other, then we say the two method sets are identical.
Given a type T, assume it is neither a pointer type nor an
interface type, for the reason
mentioned in the last section, the method set of a type T is always
a subset of the method set of type *T.
For example, the method set of the Book type shown above is a
subset of the method set of the *Book type.
Please note, non-exported method names, which start with lower-case letters, from different packages will be always viewed as two different method names, even if the two method names are the same in literal.
Method sets play an important role in the polymorphism feature of Go. About polymorphism, please read the next article (interfaces in Go) for details.
The method sets of the following types are always blank:
Methods are special functions actually. Methods are often called member functions. When a type owns a method, each value of the type will own an immutable member of a function type. The member name is the same as the method name and the type of the member is the same as the function declared with the form of the method declaration but without the receiver part.
A method call is just a call to such a member function.
For a value v, its method m can be represented
with the selector form v.m, which is a function value.
package main
import "fmt"
type Book struct {
pages int
}
func (b Book) Pages() int {
return b.pages
}
func (b *Book) SetPages(pages int) {
b.pages = pages
}
func main() {
var book Book
fmt.Printf("%T \n", book.Pages) // func() int
fmt.Printf("%T \n", (&book).SetPages) // func(int)
// &book has an implicit method.
fmt.Printf("%T \n", (&book).Pages) // func() int
// Call the three methods.
(&book).SetPages(123)
book.SetPages(123) // equivalent to the last line
fmt.Println(book.Pages()) // 123
fmt.Println((&book).Pages()) // 123
}
(Different from C language, there is not the -> operator in Go
to call methods with pointer receivers,
so (&book)->SetPages(123) is illegal in Go.)
Wait! Why does the line book.SetPages(123) in the above example compile okay?
After all, the method SetPages is not declared for the Book type.
One one hand, this can be viewed as a syntactic sugar to make programming convenient.
This sugar only works for addressable value receivers.
Compiler will implicitly take the address of the addressable value book
when it is passed as the receiver argument of a SetPages method call.
On the other hand, we should also think aBookExpression.SetPages is always a legal selector (from the syntax view),
even if the expression aBookExpression is evaluated as an unaddressable Book value,
for which case, the selector aBookExpression.SetPages is invalid (but legal).
As above just mentioned, when a method is declared for a type,
each value of the type will own a member function.
Zero values are not exceptions, whether or not the zero values
of the types are represented by nil.
package main
type StringSet map[string]struct{}
func (ss StringSet) Has(key string) bool {
// Never panic here, even if ss is nil.
_, present := ss[key]
return present
}
type Age int
func (age *Age) IsNil() bool {
return age == nil
}
func (age *Age) Increase() {
*age++ // If age is a nil pointer, then
// dereferencing it will panic.
}
func main() {
_ = (StringSet(nil)).Has // will not panic
_ = ((*Age)(nil)).IsNil // will not panic
_ = ((*Age)(nil)).Increase // will not panic
_ = (StringSet(nil)).Has("key") // will not panic
_ = ((*Age)(nil)).IsNil() // will not panic
// This following line will panic. But the
// panic is not caused by invoking the method.
// It is caused by the nil pointer dereference
// within the method body.
((*Age)(nil)).Increase()
}
Same as general function arguments, the receiver arguments are also passed by copy. So, the modifications on the direct part of a receiver argument in a method call will not be reflected to the outside of the method.
package main
import "fmt"
type Book struct {
pages int
}
func (b Book) SetPages(pages int) {
b.pages = pages
}
func main() {
var b Book
b.SetPages(123)
fmt.Println(b.pages) // 0
}
Another example:
package main
import "fmt"
type Book struct {
pages int
}
type Books []Book
func (books Books) Modify() {
// Modifications on the underlying part of
// the receiver will be reflected to outside
// of the method.
books[0].pages = 500
// Modifications on the direct part of the
// receiver will not be reflected to outside
// of the method.
books = append(books, Book{789})
}
func main() {
var books = Books{{123}, {456}}
books.Modify()
fmt.Println(books) // [{500} {456}]
}
Some off topic, if the two lines in the orders of the above
Modify method are exchanged, then both of the modifications
will not be reflected to outside of the method body.
func (books Books) Modify() {
books = append(books, Book{789})
books[0].pages = 500
}
func main() {
var books = Books{{123}, {456}}
books.Modify()
fmt.Println(books) // [{123} {456}]
}
The reason here is that the append call will allocate a new
memory block to store the elements of the copy of the passed slice receiver argument.
The allocation will not reflect to the passed slice receiver argument itself.
func (books *Books) Modify() {
*books = append(*books, Book{789})
(*books)[0].pages = 500
}
func main() {
var books = Books{{123}, {456}}
books.Modify()
fmt.Println(books) // [{500} {456} {789}]
}
At compile time, compilers will normalize each method value expression, by changing implicit address taking and pointer dereference operations into explicit ones in that method value expression.
Assumev is a value of type T
and v.m is a legal method value expression,
m is a method explicitly declared for type *T,
then compilers will normalize it as (&v).m;
m is a method explicitly declared for type T,
then the method value expression v.m is already normalized.
p is a value of type *T
and p.m is a legal method value expression,
m is a method explicitly declared for type T,
then compilers will normalize it as (*p).m;
m is a method explicitly declared for type *T,
then the method value expression p.m is already normalized.
Promoted method value Normalization will be explained in the following type embedding article.
Assume v.m is a normalized method value expression,
at run time, when the method value v.m is evaluated,
the receiver argument v is evaluated and
a copy of the evaluation result is saved and used in later calls to the method value.
b.Pages is already normalized.
At run time, a copy of the receiver argument b is saved.
The copy is the same as Book{pages: 123},
the subsequent modification of value b has no effects on this copy.
That is why the call f1() prints 123.
p.Pages is normalized as (*p).Pages at compile time.
At run time, the receiver argument *p is evaluated to the current b value,
which is Book{pages: 123}.
A copy of the evaluation result is saved and used in later calls of the method value,
that is why the call f2() also prints 123.
p.Pages2 is already normalized.
At run time, a copy of the receiver argument p is saved.
The saved value is the address of the value b,
thus any changes to b will be reflected through dereferencing of the saved value,
that is why the call g1() prints 789.
b.Pages2 is normalized as (&b).Pages2 at compile time.
At run time, a copy of the evaluation result of &b is saved.
The saved value is the address of the value b,
thus any changes to b will be reflected through dereferencing of the saved value,
that is why the call g2() prints 789.
package main
import "fmt"
type Book struct {
pages int
}
func (b Book) Pages() int {
return b.pages
}
func (b *Book) Pages2() int {
return (*b).Pages()
}
func main() {
var b = Book{pages: 123}
var p = &b
var f1 = b.Pages
var f2 = p.Pages
var g1 = p.Pages2
var g2 = b.Pages2
b.pages = 789
fmt.Println(f1()) // 123
fmt.Println(f2()) // 123
fmt.Println(g1()) // 789
fmt.Println(g2()) // 789
}
MyInt,
the defined type Age has not an IsOdd method.
package main
type MyInt int
func (mi MyInt) IsOdd() bool {
return mi%2 == 1
}
type Age MyInt
func main() {
var x MyInt = 3
_ = x.IsOdd() // okay
var y Age = 36
// _ = y.IsOdd() // error: y.IsOdd undefined
_ = y
}
Firstly, from the last section, we know that sometimes we must declare methods with pointer receivers.
In fact, we can always declare methods with pointer receivers without any logic problems. It is just a matter of program performance that sometimes it is better to declare methods with value receivers.
sync standard package
should not be copied, so declaring methods with value receivers for
struct types which embedding
the types in the sync standard package is problematic.
If it is hard to make a decision whether a method should use a pointer receiver or a value receiver, then just choose the pointer receiver way.
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reflect standard package.sync standard package.sync/atomic standard package.