Methods and Interfaces Part1
1.Methods
Go does not have classes. However, you can define methods on types.
A method is a function with a special receiver argument.
The receiver appears in its own argument list between the func keyword and the method name.
In this example, the Abs method has a receiver of type Vertex named v.
package main
import (
"fmt"
"math"
)
type Vertex struct {
X, Y float64
}
func (v Vertex) Abs() float64 {
return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func main() {
v := Vertex{3, 4}
fmt.Println(v.Abs())
}
2.Methods are functions
Remember: a method is just a function with a receiver argument.
Here's Abs written as a regular function with no change in functionality.
3.Methods continued
You can declare a method on non-struct types, too.
In this example we see a numeric type MyFloat with an Abs method.
You can only declare a method with a receiver whose type is defined in the same package as the method. You cannot declare a method with a receiver whose type is defined in another package (which includes the built-in types such as int).
package main
import (
"fmt"
"math"
)
type MyFloat float64
func (f MyFloat) Abs() float64 {
if f < 0 {
return float64(-f)
}
return float64(f)
}
func main() {
f := MyFloat(-math.Sqrt2)
fmt.Println(f.Abs())
}
4.Pointer receivers
You can declare methods with pointer receivers.
This means the receiver type has the literal syntax *T for some type T. (Also, T cannot itself be a pointer such as *int.)
For example, the Scale method here is defined on *Vertex.
Methods with pointer receivers can modify the value to which the receiver points (as Scale does here). Since methods often need to modify their receiver, pointer receivers are more common than value receivers.
Try removing the * from the declaration of the Scale function on line 16 and observe how the program's behavior changes.
With a value receiver, the Scale method operates on a copy of the original Vertex value. (This is the same behavior as for any other function argument.) The Scale method must have a pointer receiver to change the Vertex value declared in the main function.
package main
import (
"fmt"
"math"
)
type Vertex struct {
X, Y float64
}
func (v Vertex) Abs() float64 {
return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func (v *Vertex) Scale(f float64) {
v.X = v.X * f
v.Y = v.Y * f
}
func main() {
v := Vertex{3, 4}
v.Scale(10)
fmt.Println(v.Abs())
}
5.Methods and pointer indirection
Comparing the previous two programs, you might notice that functions with a pointer argument must take a pointer:
var v Vertex ScaleFunc(v, 5) // Compile error! ScaleFunc(&v, 5) // OK
while methods with pointer receivers take either a value or a pointer as the receiver when they are called:
var v Vertex v.Scale(5) // OK p := &v p.Scale(10) // OK
For the statement v.Scale(5), even though v is a value and not a pointer, the method with the pointer receiver is called automatically. That is, as a convenience, Go interprets the statement v.Scale(5) as (&v).Scale(5) since the Scale method has a pointer receiver.
package main
import "fmt"
type Vertex struct {
X, Y float64
}
func (v *Vertex) Scale(f float64) {
v.X = v.X * f
v.Y = v.Y * f
}
func ScaleFunc(v *Vertex, f float64) {
v.X = v.X * f
v.Y = v.Y * f
}
func main() {
v := Vertex{3, 4}
v.Scale(2)
ScaleFunc(&v, 10)
p := &Vertex{4, 3}
p.Scale(3)
ScaleFunc(p, 8)
fmt.Println(v, p)
}
6.Methods and pointer indirection (2)
The equivalent thing happens in the reverse direction.
Functions that take a value argument must take a value of that specific type:
var v Vertex fmt.Println(AbsFunc(v)) // OK fmt.Println(AbsFunc(&v)) // Compile error!
while methods with value receivers take either a value or a pointer as the receiver when they are called:
var v Vertex fmt.Println(v.Abs()) // OK p := &v fmt.Println(p.Abs()) // OK
In this case, the method call p.Abs() is interpreted as (*p).Abs().
package main
import (
"fmt"
"math"
)
type Vertex struct {
X, Y float64
}
func (v Vertex) Abs() float64 {
return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func AbsFunc(v Vertex) float64 {
return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func main() {
v := Vertex{3, 4}
fmt.Println(v.Abs())
fmt.Println(AbsFunc(v))
p := &Vertex{4, 3}
fmt.Println(p.Abs())
fmt.Println(AbsFunc(*p))
}
7.Choosing a value or pointer receiver
There are two reasons to use a pointer receiver.
The first is so that the method can modify the value that its receiver points to.
The second is to avoid copying the value on each method call. This can be more efficient if the receiver is a large struct, for example.
In this example, both Scale and Abs are methods with receiver type *Vertex, even though the Abs method needn't modify its receiver.
In general, all methods on a given type should have either value or pointer receivers, but not a mixture of both. (We'll see why over the next few pages.)
package main
import (
"fmt"
"math"
)
type Vertex struct {
X, Y float64
}
func (v *Vertex) Scale(f float64) {
v.X = v.X * f
v.Y = v.Y * f
}
func (v *Vertex) Abs() float64 {
return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func main() {
v := &Vertex{3, 4}
fmt.Printf("Before scaling: %+v, Abs: %v\n", v, v.Abs())
v.Scale(5)
fmt.Printf("After scaling: %+v, Abs: %v\n", v, v.Abs())
}
8.Interfaces
An interface type is defined as a set of method signatures.
A value of interface type can hold any value that implements those methods.
Note: There is an error in the example code on line 22. Vertex (the value type) doesn't implement Abser because the Abs method is defined only on *Vertex (the pointer type).
9.Interfaces are implemented implicitly
A type implements an interface by implementing its methods. There is no explicit declaration of intent, no "implements" keyword.
Implicit interfaces decouple the definition of an interface from its implementation, which could then appear in any package without prearrangement.
