go - 将方法转换为 func var

Question

为什么将方法分配给函数处理程序有效。

短链接:https://play.golang.org/p/UEYGCpMgyV6

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 Abs() float64 {
    return math.Sqrt(10)
}

func AbsFloat(f float64) float64 {
    return math.Sqrt(f)
}

func main() {
    v := Vertex{3, 4}

    // Assigning Abs() to func handler.
    var absFunc func() float64 = Abs
    fmt.Println(absFunc())

    // Wrong type as expected.
    //absFunc = AbsFloat

    // Assigning method to func handler works, why ?
    absFunc = v.Abs

    // Changing receiver args. Useful for unittesting but why/how this works ?
    v.X = 1
    v.Y = 1
    fmt.Println(absFunc())
}

它是 Golang 类型检查“功能” func()float64 == func(T)float64 还是其他？请帮助理解这个概念。

Answer 1

v.Abs 是一个 method value ，并引用规范:

The method value x.M is a function value that is callable with the same arguments as a method call of x.M.

方法值具有与没有接收者的方法相同的参数和结果类型的函数类型。