There is no explicit declaration that a type implements an interface. Infact, in Go there doesn’t exist any “implements” keyword similar to Java. A type implements an interface if it implements all the methods of the interface.
Let’s see an example
package main
import "fmt"
type animal interface {
breathe()
walk()
}
type lion struct {
age int
}
func (l lion) breathe() {
fmt.Println("Lion breathes")
}
func (l lion) walk() {
fmt.Println("Lion walk")
}
func main() {
var a animal
a = lion{age: 10}
a.breathe()
a.walk()
}Output
Lion breathes
Lion walkThere is no explicit declaration which says that lion struct implements the animal interface. During compilation, go notices that lion struct implements all methods of animal interface hence it is allowed.Any other type which implements all methods of the animal interface becomes of that interface type.
This holds true even when the interface and the type which defines all methods of an interface are in different packages
Let’s see a more complex example of another type implementing the animal interfaceIf we define a dog struct and it implements the breathe and walk method then it will also be an animal. Let’s see an example
package main
import "fmt"
type animal interface {
breathe()
walk()
}
type lion struct {
age int
}
func (l lion) breathe() {
fmt.Println("Lion breathes")
}
func (l lion) walk() {
fmt.Println("Lion walk")
}
type dog struct {
age int
}
func (l dog) breathe() {
fmt.Println("Dog breathes")
}
func (l dog) walk() {
fmt.Println("Dog walk")
}
func main() {
var a animal
a = lion{age: 10}
a.breathe()
a.walk()
a = dog{age: 5}
a.breathe()
a.walk()
}Output
Lion breathes
Lion walk
Dog breathes
Dog walkBoth lion and dog implement the breathe and walk method hence they are of animal type and can correctly be assigned to a variable of interface type.