英文:
Golang method receiver
问题
根据这个问题，Golang会生成type-receiver method和point-receiver method，这意味着下面的代码将正常工作，但值会意外地改变。
func (test *Test) modify() {
test.a++
}
func main() {
test := Test{10}
fmt.Println(test)
test.modify()
fmt.Println(test)
}
我认为这对我来说是可以接受的。但是当这与接口混合使用时，问题就出现了。
type Modifiable interface {
modify()
}
type Test struct {
a int
}
func (test *Test) modify() {
test.a++
}
func main() {
test := Test{10}
fmt.Println(test)
test.modify()
fmt.Println(test)
var a Modifiable
a = test
}
它报错说：
"Test does not implement Modifiable (modify method has pointer receiver)"
为什么会发生这种情况？
Golang实际上是如何处理方法调用的？
英文:
According this question, golang will generate both type-receiver method and point-receiver method, which means the code below will work correctly and the value will change unexpectedly.
func (test *Test) modify() {
test.a++
}
func main() {
test := Test{10}
fmt.Println(test)
test.modify()
fmt.Println(test)
}
I think it's acceptable to me. But when this mixes with interface, thing goes wrong.
type Modifiable interface {
modify()
}
type Test struct {
a int
}
func (test *Test) modify() {
test.a++
}
func main() {
test := Test{10}
fmt.Println(test)
test.modify()
fmt.Println(test)
var a Modifiable
a = test
}
it said:
Test does not implement Modifiable (modify method has pointer receiver)
Why will this happen ?
And how golang actually handle method call ?
答案1
得分: 1
如果你想使用具有指针接收器的方法，这意味着你必须传递地址值。
这是一个例子：
package main
import "fmt"
type Modifiable interface {
modify()
}
type Test struct {
a int
}
func (test *Test) modify() {
test.a++
}
func main() {
test := Test{10}
fmt.Println(test)
test.modify()
fmt.Println(test)
var a Modifiable
a = &test
a.modify()
fmt.Println(a)
}
总结一下，当你在方法中创建一个指针接收器时，接口将接受地址值。
英文:
if you wanted to use a method that has pointer receiver. It means you have to pass the address value.
here is the example :
package main
import "fmt"
type Modifiable interface {
modify()
}
type Test struct {
a int
}
func (test *Test) modify() {
test.a++
}
func main() {
test := Test{10}
fmt.Println(test)
test.modify()
fmt.Println(test)
var a Modifiable
a = &test
a.modify()
fmt.Println(a)
}
In conclusion an interface will accept the address value whenever you create a pointer receiver in the method.
答案2
得分: 1
当你说：
func (test *Test) modify() {
test.a++
}
这意味着类型 *Test 实现了接口 Modifiable，也就是 指向 Test 的指针
而
func (test Test) modify() {
test.a++
}
意味着类型 Test 实现了接口
结论是：类型和指向该类型的指针是两种不同的类型。
英文:
When you said:
func (test *Test) modify() {
test.a++
}
It means the interface Modifiable is implemented by the type *Test aka the Pointer to Test
Where as
func (test Test) modify() {
test.a++
}
means that the interface is implemented by the type Test
Conclusion is: A type and a pointer to that type are 2 different types.
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