It works when I changed the code as your suggested. That's great, thanks.
And I'm still a little confused here, you know when we use the struct
method directly, it is only when the function is called that the type of
receiver determines whether the passed struct is a pointer or a copied
value. But when using a function pointer, why does it decide whether to
bind a pointer or a copied value at the time of assignment, but not at the
time of function called?
It seems that these two behaviors are not consistent. Is there any benefit
to doing so? I didn't find much information on this topic on Google. Is
there any extended reading on this topic?
On Tuesday, August 10, 2021 at 12:07:26 PM UTC-7 bse...@computer.org wrote:
> On Tue, Aug 10, 2021 at 12:01 PM E Z <lege...@gmail.com> wrote:
>
>> I feel confused when I use the function pointer which point to the struct
>> method. Here is the test code:
>>
>> /***
>> package main
>>
>> type Zoo struct {
>> Animal string
>> }
>>
>> func (z Zoo) Display(){
>> fmt.Printf("Current animal is:%s\n", z.Animal)
>> }
>>
>
> This method has a value receiver. When pf is assigned to gz.Display, it is
> assigned with its receiver, which is a copy of gz.
>
> Change the method to func (z *Zoo) Display(), and it will work as you
> expect.
>
>
>>
>> func main(){
>> gz := &Zoo{
>> Animal: "Monkey",
>> }
>>
>> pf := gz.Display
>> pf() //display "Current animal
>> is Monkey"
>> gz.Animal="Tiger"
>> pf() //display "Current animal
>> is Monkey"
>> gz.Display() //display "Current animal is
>> Tiger"
>> }
>> ***/
>> As the code is shown above, even though I changed the value of the
>> member field of the Zoo instance gz, the function pointer call that
>> followed still prints the unmodified value.
>>
>> Can someone help explain why? In my opinion, The function pointer pf
>> should bind to the instance gz and its method(note that gz here is a
>> pointer variable), if so anytime I change the value of the variable gz,
>> pf should reflect the changes.
>>
>> Thanks,
>> Ethan
>>
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>>
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