On 2020-04-21 10:09, Thomas S wrote:
>> But most of the time for getting a copy of a struct, I do something like
>> this :
>>
>> func (re Regexp) Copy() *Regexp {
>> 119 return &re
>> 120 }
If you are just trying to shallow copy a struct then I find this pointer usage
distasteful.
Far mor
I did not know this stylistic guideline.
Thank you Ian !
Le mardi 21 avril 2020 00:42:39 UTC+2, Ian Lance Taylor a écrit :
>
> On Mon, Apr 20, 2020 at 1:56 PM Thomas S >
> wrote:
> >
> > In regexp package, the "copy" function is implemented like this :
> >
> > func (re *Regexp) Copy() *Regex
On Mon, Apr 20, 2020 at 1:56 PM Thomas S wrote:
>
> In regexp package, the "copy" function is implemented like this :
>
> func (re *Regexp) Copy() *Regexp {
>118 re2 := *re
>119 return &re2
>120 }
>
>
> But most of the time for getting a copy of a struct, I do something like this
Hello,
In regexp package, the "copy" function is implemented like this :
func (re *Regexp) Copy() *Regexp { 118re2 := *re 119return
&re2 120 }
But most of the time for getting a copy of a struct, I do something like
this :
func (re Regexp) Copy() *Regexp { 119
