再看go的interface程式碼示例
阿新 • • 發佈：2018-11-10
程式碼：
package main import "fmt" type Base interface { Input() int } type Dog struct { } func (p Dog) Input() int { fmt.Println("call input dog") return 100 } func main() { handler := func() Base { return Dog{} } fmt.Printf("type: %T, %v\n", handler(), handler().Input()) }
結果：
call input dog
type: main.Dog, 100
程式碼：
package main import "fmt" type Base interface { Input() int } type Dog struct { } func (p Dog) Input() int { fmt.Println("call input dog") return 100 } func main() { handler := func() Base { return &Dog{} } fmt.Printf("type: %T, %v\n", handler(), handler().Input()) }
結果：
call input dog
type: *main.Dog, 100
程式碼：
package main import "fmt" type Base interface { Input() int } type Dog struct { } func (p *Dog) Input() int { fmt.Println("call input dog") return 100 } func main() { handler := func() Base { return &Dog{} } fmt.Printf("type: %T, %v\n", handler(), handler().Input()) }
結果：
call input dog
type: *main.Dog, 100
程式碼：
package main
import "fmt"
type Base interface {
Input() int
}
type Dog struct {
}
func (p *Dog) Input() int {
fmt.Println("call input dog")
return 100
}
func main() {
handler := func() Base { return Dog{} }
fmt.Printf("type: %T, %v\n", handler(), handler().Input())
}
結果：
# command-line-arguments
./a.go:17:40: cannot use Dog literal (type Dog) as type Base in return argument:
Dog does not implement Base (Input method has pointer receiver)
好好理解下。