You are probably thinking about code like this:
var f2 = *f1
Which will make a copy, although not because `f1` is dereferenced, but
because `=` was called on a value.
Dereferencing a pointer gives a reference to the same value, taking
address of the same value will produce a pointer to the same value. So
in expression like this:
var f2 = &(*f1)
`&` and `*` cancel each other out and it can be simplified to:
var f2 = f1
Which means "make a copy of pointer". If you take the address of `f1`
and `f2` you'll see that those are indeed different pointers.
If you're thinking about something like `f(*f1)`, then I believe the
function call will make a copy because arguments are passed by values.
`f(f1)` will make a copy too, but a copy of pointer.
Does that make sense?
сб, 25 июл. 2020 г. в 11:09, chri...@surlykke.dk <christ...@surlykke.dk>:
>
> When running this program:
>
> package main
>
> import (
> "fmt"
> )
>
> type Foo struct {
> i int
> }
>
> func main() {
> var f1 = &Foo{i: 0}
> var f2 = &(*f1)
> f2.i = 1
> fmt.Println(f1, f2)
> }
>
> it yields:
>
> &{1} &{1}
>
> (https://play.golang.org/p/qKtURokUCEW)
>
> I (naively) assumed that the expression
>
> &(*f1)
>
> would, first, create a copy of *f1, and then a pointer to that copy, but
> evidently f2 becomes a pointer to the same struct as f1. Is this something I
> should have deduced from the language spec?
>
> best regards Christian Surlykke
>
>
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