I understand the worry but the more I think about it, the less I think it
would end up being a real issue.
Taking the example of maps... two different maps that have not the same
location in memory are never equal, especially since modifying one does not
modify the other. It's easily explained.
For slices, explaining that it's a view of an array, and views of different
arrays are not considered equals, it would also be easily explained to a
beginning programmer.
Well it's merely food for thoughts. (but it would solve later issues in the
usability of maps)
On Thursday, June 30, 2016 at 6:28:47 PM UTC+2, Ian Lance Taylor wrote:
>
> On Thu, Jun 30, 2016 at 9:24 AM, Chad <send...@gmail.com <javascript:>>
> wrote:
> > It's a view in another array.
> > Why should they be equal? Unless the second slice is constructed by
> > subslicing the other as such `[:]`, the slices are different.
> >
> > If I were to access one of the slice backing array and mutate it, I
> wouldn't
> > expect the slices to be equal anymore.
> >
> > The argument that it would be surprising is perhaps flawed because it
> stems
> > from a misconception.
> > A slice is a dynamic view. For two slices to be equal it makes sense to
> be
> > looking at the same thing at all time.
> >
> > Taken mathematically, if we describe an indirection as a mathematical
> > function, a slice is such a function. Two functions are equals if they
> are
> > equals everywhere they are defined. A slice is equal to another slice if
> the
> > backign arrays always have the same values, whatever mutation occurs on
> any
> > of them.
> >
> > Comparing the snapshot view of a slice would then different. The
> behaviour
> > seems consistent to me.
>
> This discussion is a good example of why the language does not define
> the behavior. There are solid arguments on both sides.
>
> Ian
>
>
> > On Thursday, June 30, 2016 at 4:54:59 PM UTC+2, Paul Borman wrote:
> >>
> >> It would be very surprising to people to use a slice as a key to map,
> say
> >> []int{1,2} and then find that using another []int{1,2} does not find
> that
> >> entry. While I think it would be nice to have == on slices, I must
> agree
> >> with Ian and authors that it is better left unsaid.
> >>
> >> -Paul
> >>
> >> On Thu, Jun 30, 2016 at 1:09 AM, Chad <send...@gmail.com> wrote:
> >>>
> >>>
> >>>
> >>> On Thursday, June 30, 2016 at 8:32:58 AM UTC+2, Viktor Kojouharov
> wrote:
> >>>>
> >>>> This would probably introduce unnecessary confusion. People are used
> to
> >>>> the equality operator comparing values in go, as opposed to
> references. It's
> >>>> much better if the slices finally support the equality operator, even
> though
> >>>> the comparison speed will depend on the number of items in the
> slices.
> >>>>>
> >>>>>
> >>> Well, in fact, everything is a value in Go. It's just that the
> internal
> >>> structure of the standard reference types are not exposed.
> >>> The behaviour should not be very different from the comparison between
> >>> user defined structs where one or more fields are pointers.
> >>>
> >>> It also makes sense to say that two slices are equal iff they
> represent
> >>> the exact same views of the exact same underlying array.
> >>>
> >>>
> >>> --
> >>> You received this message because you are subscribed to the Google
> Groups
> >>> "golang-nuts" group.
> >>> To unsubscribe from this group and stop receiving emails from it, send
> an
> >>> email to golang-nuts...@googlegroups.com.
> >>> For more options, visit https://groups.google.com/d/optout.
> >>
> >>
> > --
> > You received this message because you are subscribed to the Google
> Groups
> > "golang-nuts" group.
> > To unsubscribe from this group and stop receiving emails from it, send
> an
> > email to golang-nuts...@googlegroups.com <javascript:>.
> > For more options, visit https://groups.google.com/d/optout.
>
--
You received this message because you are subscribed to the Google Groups
"golang-nuts" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to golang-nuts+unsubscr...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.
