Go: *Var 是 Var 的 "subclass"吗？

Question

取自 go tour:

package main

import (
    "fmt"
    "math"
)

type Abser interface {
    Abs() float64
}

func main() {
    var a Abser
    f := MyFloat(-math.Sqrt2)
    v := Vertex{3, 4}

    a = f
    a = &v

    // v == Vertex != *Vertex -> exception
    a = v
}

type MyFloat float64

func (f MyFloat) Abs() float64 {
    if f < 0 {
        return float64(-f)
    }
    return float64(f)
}

type Vertex struct {
    X, Y float64
}

func (v *Vertex) Abs() float64 {
    return math.Sqrt(v.X*v.X + v.Y*v.Y)
}

但是，当将 func (v *Vertex) Abs() float64 转换为 func (v Vertex) Abs() float64 时，代码编译:

package main

import (
    "math"
)

type Abser interface {
    Abs() float64
}

func main() {
    var a Abser
    f := MyFloat(-math.Sqrt2)
    v := Vertex{3, 4}

    a = f

    // Since *Vertex != Vertex, this shouldn't compile, should it?
    a = &v

    a = v
}

type MyFloat float64

func (f MyFloat) Abs() float64 {
    if f < 0 {
        return float64(-f)
    }
    return float64(f)
}

type Vertex struct {
    X, Y float64
}

func (v Vertex) Abs() float64 {
    return math.Sqrt(v.X*v.X + v.Y*v.Y)
}

为什么第二个例子运行？

Answer 1

类型*T 不是T 的子类，而是*T 的method set将继承T的方法:

The method set of any other type T consists of all methods declared with receiver type T. The method set of the corresponding pointer type *T is the set of all methods declared with receiver *T or T (that is, it also contains the method set of T).

因此，如果 T 符合特定接口(interface)，那么 *T 也符合。这就是为什么您可以将 *Vertex 值分配给示例中的 Abser 变量。