On Friday, October 21, 2016 at 11:37:21 PM UTC+8, Paul Borman wrote:
>
> I think you should clarify that this is because T *only* contains
> pointers. If T were:
>
> type T struct {
> i int
> p *int
> }
>
>
Ian says slice is also a reference type. This type declaration is much like
slice type:
type Slice struct {
array *internalArray
len int
cap 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 <
> golan...@googlegroups.com <javascript:>> wrote:
>
>> On 21 October 2016 at 11:15, T L <tapi...@gmail.com <javascript:>> 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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>
>
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