英文:
Methods and receivers in Go
问题
我有困难理解Go语言中的方法和接收器。假设我们有以下代码：
package main
import ("fmt"; "math")
type Circle struct {
x, y, r float64
}
func (c *Circle) area() float64 {
return math.Pi * c.r * c.r
}
func main() {
c := Circle{0, 0, 5}
fmt.Println(c.area())
}
在area函数的定义中，(c *Circle)被称为接收器，在main函数中，我们可以通过引用调用area并传递c，而不使用指针。我可以将代码编辑为以下形式，它的工作方式相同：
package main
import ("fmt"; "math")
type Circle struct {
x, y, r float64
}
func circleArea(c *Circle) float64 {
return math.Pi * c.r*c.r
}
func main() {
c := Circle{0, 0, 5}
fmt.Println(circleArea(&c))
}
现在，这只是两段代码之间的语法差异，还是在更深层次上存在一些结构上的不同？
英文:
I have problems understanding methods and receivers in Go. Let's say we have this code:
package main
import ("fmt"; "math")
type Circle struct {
x, y, r float64
}
func (c *Circle) area() float64 {
return math.Pi * c.r * c.r
}
func main() {
c := Circle{0, 0, 5}
fmt.Println(c.area())
}
(c *Circle) in the definition of the area function is said to be a receiver and in the main we can call area and pass c by reference without the use of pointers. I can edit the code to the following and it works the same way:
package main
import ("fmt"; "math")
type Circle struct {
x, y, r float64
}
func circleArea(c *Circle) float64 {
return math.Pi * c.r*c.r
}
func main() {
c := Circle{0, 0, 5}
fmt.Println(circleArea(&c))
}
Now is this just a syntactical difference between the two snippets of code or is there something structurally different going on on a deeper level?
答案1
得分: 6
这是要翻译的内容：
区别不仅仅在于语法。使用方法，你的圆形类型可以实现一个接口，但是函数却不能做到这一点：
type areaer interface {
area() float64
}
英文:
The difference isn't just syntax. With a method, your circle type could fulfill an interface, but the function doesn't let you do that:
type areaer interface {
area() float64
}
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