golang之method

這是一個建立於 的文章，其中的資訊可能已經有所發展或是發生改變。

method

Go does not have classes. However, you can define methods on types.

package mainimport (    "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 (v Vertex) ScaleValue(f float64) {    v.X = v.X * f    v.Y = v.Y * f}func main() {    v := Vertex{3, 4}    fmt.Println(v.Abs())    v.Scale(2)    fmt.Println(v.Abs())    v.ScaleValue(2)    fmt.Println(v.Abs())    p := &v    p.Scale(2)    fmt.Println(p.Abs())}

輸出如下：

5
10
10
20

 

三個注意點：

1. Methods with pointer receivers can modify the value to which the receiver points 。

2. methods with pointer receivers take either a value or a pointer as the receiver when they are called:

var v Vertexv.Scale(5)  // OKp := &vp.Scale(10) // OK

3. methods with value receivers take either a value or a pointer as the receiver when they are called:

var v Vertexfmt.Println(v.Abs()) // OKp := &vfmt.Println(p.Abs()) // OK

interface

package mainimport (    "fmt")type I interface {    M()}type T struct {    S string}func (t *T) M() {    fmt.Println(t.S)}func main() {    var i I    t := T{"hello"}    i = &t    i.M()    v := T{"world"}    i = v    i.M()}

main函數後半段代碼會報錯：cannot use v (type T) as type I in assignment: T does not implement I (M method has pointer receiver)

如果把M的receiver類型改成value：

func (t T) M() {    fmt.Println(t.S)}

main函數的代碼將正常運行。