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.
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 <javascript:>>
> 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.
>>
>>
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>
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