1
考虑下面这个简单的围棋程序函数是否必须满足函数类型的确切签名?
package main
import (
"io"
"encoding/json"
"os"
)
type MyEncoder interface {
Encode(v interface{}) error
}
type MyEncoderCreator func(io.Writer) *MyEncoder
type MyContainer struct {
Creator MyEncoderCreator
}
func main() {
container := Container{
Creator:json.NewEncoder,
}
encoder := container.Creator(os.Stdout)
encoder.Encode(map[string]string{"key":"value"})
}
这项计划失败,出现以下错误编译:
./main.go:21: cannot use json.NewEncoder (type func(io.Writer) *json.Encoder) as type MyEncoderCreator in field value
这是为什么? json.Encoder
结构具有满足接口MyEncoder
的接收器。所以json.NewEncoder
函数应该被允许分配给MyContainer.Creator
?