英文:
Calling a method with a pointer receiver by an object instead of a pointer to it?
问题
v是Vertex的一个对象,Scale是指向Vertex的指针的方法。那么为什么v.Scale(10)不是错误的,考虑到v不是指向Vertex对象的指针?谢谢。
package main
import (
"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 main() {
v := Vertex{3, 4}
v.Scale(10)
fmt.Println(v.Abs())
}
英文:
v is an object of Vertex, and Scale is a method for a pointer to Vertex. Then why is v.Scale(10) not wrong, given that v isn't a pointer to a Vertex object? Thanks.
package main
import (
"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 main() {
v := Vertex{3, 4}
v.Scale(10)
fmt.Println(v.Abs())
}
答案1
得分: 11
[规范:调用:]
如果方法集(类型的)
x包含m,并且参数列表可以赋值给m的参数列表,那么方法调用x.m()是有效的。如果x是可寻址的,并且&x的方法集包含m,x.m()就是(&x).m()的简写。
编译器看到 Scale() 有一个指针接收器,并且 v 是可寻址的(因为它是一个局部变量),所以 v.Scale(10) 将被解释为 (&v).Scale(10)。
这只是规范提供的许多便利之一,以使源代码保持简洁。
英文:
> A method call x.m() is valid if the method set of (the type of) x contains m and the argument list can be assigned to the parameter list of m. If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m().
The compiler sees that Scale() has a pointer receiver and also that v is addressable (as it is a local variable), so v.Scale(10) will be interpreted as (&v).Scale(10).
This is just one of the many conveniences the spec offers you so the source code can remain clean.
答案2
得分: 4
这是Go语言的自动解引用:
从https://golang.org/ref/spec#Method_values:
与选择器一样,使用指针引用值接收器的非接口方法将自动解引用该指针:pt.Mv 等同于 (*pt).Mv。
英文:
It's the Go automatic dereferencing:
From https://golang.org/ref/spec#Method_values:
>As with selectors, a reference to a non-interface method with a value receiver using a pointer will automatically dereference that pointer: pt.Mv is equivalent to (*pt).Mv.
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