pointers - GO - 隐式方法如何工作？

Question

In GO, I learnt that,

1)

Programmer can only define methods on named types(X) or pointer(*X) to named types

2)

An explicit method definition for type X implicitly defines the same method for type *X and vice versa, so, my understanding is, If I declare,

func (c Cat) foo(){
  //do stuff_
} 

and declare,

func (c *Cat) foo(){
  // do stuff_
}

then GO compiler gives,

Compile error: method re-declared

which indicates that, pointer method is implicitly defined and vice versa

---

With the given named type(Cat),

type Cat struct{
  Name string
  Age int
  Children []Cat
  Mother *Cat
} 

---

Scenario 1

Method(foo) defined on a named type(Cat), by programmer,

func (c Cat) foo(){
   // do stuff....
}

that implicitly defines method(foo) on pointer(*Cat) to named type, by GO compiler, that looks like,

func (c *Cat) foo(){
   // do stuff....
}

On creating variables of named type(Cat)

var c Cat
var p *Cat = &c

c.foo() has the method defined by programmer.

Question 1:

On invoking p.foo(), does implicit pointer method receive the pointer(p)?

---

Scenario 2

Method(foo) defined on a pointer(*Cat) to named type, by the programmer,

func (c *Cat) foo(){
   // do stuff....
  }

that implicitly defines method(foo) on named type(Cat), by the GO compiler, that looks like,

func (c Cat) foo(){
   // do stuff....
  }

On creating variables of named type(Cat)

var c Cat
var p *Cat = &c

p.foo() has method defined by programmer(above).

Question 2:

On invoking c.foo(), does the implicit non-pointer method receive the value c?

Answer 1

1. 类型 X 的显式方法定义隐式定义类型 *X 的相同方法，反之亦然。

这是不正确的。方法不是隐式定义的。编译器为您做的唯一一件事就是将 c.foo() 隐式替换为 (*c).foo() 或 c.foo() 和 (&c).foo() 为了方便。查看tour of Go

1. 根据您的显式声明，方法的接收者是 T 类型或 *T 类型。