Why implicit non-pointer methods not satisfy interface?

Question

Assuming we have an understanding that,

For explicit method definition for type X, GO compiler implicitly defines the same method for type *X and vice versa, if I declare,

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

and declare,

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

then GO compiler gives error,

Compile error: method re-declared

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

---

In the below code,

package main

type X interface{
  foo();
  bar();
}

type Cat struct{

}

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

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

func main() {
  var c Cat
  var p *Cat
  var x X

  x = p // OK; *Cat has explicit method bar() and implicit method foo()
  x = c //compile error: Cat has explicit method foo() and implicit method bar()

}

---

GO compiler gives error,

cannot use c (type Cat) as type X in assignment:
    Cat does not implement X (bar method has pointer receiver)

at x = c, because, implicit pointer methods satisfy interfaces, but implicit non-pointer methods do not.

Question:

Why implicit non-pointer methods do not satisfy interfaces?

Answer 1

Let's look into the language specification:

A type may have a method set associated with it. The method set of an interface type is its interface. 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).

In your example, the method set of the interface type x is [foo(), bar()]. The method set of the type Cat is [foo()], and the method set of the type *Cat is [foo()] + [bar()] = [foo(), bar()].

This explains, why variable p satisfies the interface x, but variable c doesn't.

Answer 2

Method set

Following the spec:

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

Method set definition sounds weird until you follow addressable and not addressable types concept.

Addressable and not addressable types

It is possible to call a pointer receiver method on a value if the value is of addressable type.

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.

As with method calls, a reference to a non-interface method with a pointer receiver using an addressable value will automatically take the address of that value: t.Mp is equivalent to (&t).Mp.

It is ok to call pointer receiver methods on values till you are dealing with addressable types (struct is addressable):

type Cat struct {}

func (c *Cat) bar() string { return "Mew" }

func main() {
    var c Cat
    c.bar()
}

Variables of interface type are not addressable

But not all Go types are addressable. Also variables referenced through interfaces are not addressable.

It is impossible to call pointer receiver on values of not addressable types:

type X interface {
    bar() string
}

type Cat struct{}

func (c *Cat) bar() string { return "Mew" }

/* Note `cat` variable is not a `struct` type value but
   it is type of `X` interface therefor it is not addressable. */
func CatBar(cat X) { 
    fmt.Print(cat.bar())
}

func main() {
    var c Cat
    CatBar(c)
}

So with the following error Go runtime prevents segment fault:

cannot use c (type Cat) as type X in assignment: Cat does not implement X (bar method has pointer receiver)

Answer 3

Add a little to dev.bmax's answer.

type Cat struct{
}

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

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

you can do

var c cat
c.bar() // ok to call bar(), since c is a variable.

but not

cat{}.bar() // not ok to call bar(), c is not a variable.

It's legal to call a *T method on an argument of type T so long as the argument is a variable; the compiler implicitly takes its address. But this is mere syntactic sugar: a value of type T does not posses all methods that a *T pointer does, and as a result it might satisfy fewer interfaces.

On the other hand, you can always call foo() with Cat or *Cat.

Answer 4

How about this?

package main

import (
    "fmt"
)

type Growler interface{
    Growl() bool
}

type Cat struct{
    Name string
    Age int
} 

// *Cat is good for both objects and "references" (pointers to objects)
func (c *Cat) Speak() bool{
    fmt.Println("Meow!")
        return true
}

func (c *Cat) Growl() bool{
    fmt.Println("Grrr!")
    return true
}

func main() {
    var felix Cat // is not a pointer
    felix.Speak() // works :-)
    felix.Growl() // works :-)

    var ginger *Cat = new(Cat) 
    ginger.Speak() // works :-)
    ginger.Growl() // works :-)
}