通过对象而不是指向它的指针调用带有指针接收器的方法? [英] Calling a method with a pointer receiver by an object instead of a pointer to it?

问题描述

v 是Vertex 的对象，Scale 是指向Vertex 的指针的方法.那么为什么 v.Scale(10) 没有错，因为 v 不是指向 Vertex 对象的指针?谢谢.

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())
}

推荐答案

Spec: Calls:

如果方法集xm() 是有效的> of (the type of) x 包含 m 并且参数列表可以分配给 m 的参数列表.如果 x 是 addressable 和 &x 的方法集包含 m，xm() 是 (&x).m() 的简写>.

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().

编译器看到 Scale() 有一个指针接收器，而且 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).

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).

这只是规范为您提供的众多便利之一，因此源代码可以保持干净.

This is just one of the many conveniences the spec offers you so the source code can remain clean.

这篇关于通过对象而不是指向它的指针调用带有指针接收器的方法?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！