I think you should clarify that this is because T *only* contains
pointers. If T were:
type T struct {
i int
p *int
}
then it would suddenly become non-reference type, as defined in this
thread, as a change to i will not be noticed by other copies of a given T.
-Paul
On Fri, Oct 21, 2016 at 8:20 AM, 'Alan Donovan' via golang-nuts <
golang-nuts@googlegroups.com> wrote:
> On 21 October 2016 at 11:15, T L <tapir....@gmail.com> wrote:
>>
>> On Friday, October 21, 2016 at 10:01:43 PM UTC+8, Ian Lance Taylor wrote:
>>>
>>> On Fri, Oct 21, 2016 at 6:52 AM, Henrik Johansson <dahan...@gmail.com>
>>> wrote:
>>> > The confusion I have had is rather with nilability.
>>> > A channel can be nil even though it is not explicitly a pointer.
>>>
>>> It's a basic design decision in Go that every type has a zero value.
>>> For the "reference types" (pointer, channel, map, slice, interface)
>>> that zero value is named "nil".
>>>
>>
>> I have a question, should the following type be called reference type?
>>
>> type T struct {
>> p *int
>> }
>>
>
> By the definition I gave, yes, because an instance of T contains a
> reference to an int variable. All copies of a given T value share the same
> int variable, and a change to that variable by any one will be observed by
> all the others.
>
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