On Friday, February 2, 2018 at 11:39:44 AM UTC-5, Axel Wagner wrote:
>
> I did end up asking git, why this was done and it gave this answer
> <https://github.com/golang/go/commit/2d9378c7f6dfbbe82d1bbd806093c2dfe57d7e17>,
>
> which links to this discussion <https://github.com/golang/go/issues/6398>.
>
> Hope that helps
>
Thanks for the links.
I still think allowing untyped expressions in switch would make programming
more convenient,
though it will bring much complexities to the implementation.
>
> On Fri, Feb 2, 2018 at 5:18 PM, <di...@veryhaha.com <javascript:>> wrote:
>
>>
>>
>> On Friday, February 2, 2018 at 10:45:58 AM UTC-5, Axel Wagner wrote:
>>>
>>> From the spec:
>>>
>>>
>>>> If the switch expression evaluates to an untyped constant, it is first
>>>> converted to its default type; if it is an untyped boolean value, it is
>>>> first converted to type bool. The predeclared untyped value nil cannot be
>>>> used as a switch expression.
>>>
>>>
>>> The current behavior is still as specified. Given that there is an
>>> explicit statement about how this is handled makes it less convincing,
>>> however, that the spec would be more complicated with a different behavior.
>>> So you might still file a language change request, if you want.
>>>
>>
>> but, why design as such? Isn't untyped bool more convenient?
>>
>>
>>>
>>> On Fri, Feb 2, 2018 at 4:39 PM, <di...@veryhaha.com> wrote:
>>>
>>>>
>>>>
>>>> On Friday, February 2, 2018 at 10:37:24 AM UTC-5, di...@veryhaha.com
>>>> wrote:
>>>>
>>>>>
>>>>>
>>>>> On Friday, February 2, 2018 at 10:27:04 AM UTC-5, Ian Lance Taylor
>>>>> wrote:
>>>>>>
>>>>>> On Fri, Feb 2, 2018 at 7:10 AM, <di...@veryhaha.com> wrote:
>>>>>> >
>>>>>> > Why not make it untyped?
>>>>>> >
>>>>>> > package main
>>>>>> >
>>>>>> > type T bool
>>>>>> >
>>>>>> > func f() T {return T(false)}
>>>>>> >
>>>>>> > func main() {
>>>>>> > switch {
>>>>>> > case f(): // invalid case f() in switch (mismatched types T and
>>>>>> bool)
>>>>>> > }
>>>>>> > }
>>>>>>
>>>>>> The current language spec says that omitting the switch expression is
>>>>>> equivalent to writing `true`. If you actually write `true`, then,
>>>>>> following the usual rules for untyped constant expressions, it will
>>>>>> receive the type `bool`, and you would get the same mismatched type
>>>>>> error. You will see a similar case if you write `switch 0` and try
>>>>>> to
>>>>>> compare with a named version of `int`. So the compiler is following
>>>>>> the language spec as currently written.
>>>>>>
>>>>>> We could change the language spec to say that omitting the switch
>>>>>> expression is not equivalent to writing `true`, but instead, as you
>>>>>> suggest, compares each case to an untyped `true` value. I think the
>>>>>> main argument against that is that it makes the spec slightly more
>>>>>> complicated while bringing very little benefit. You could write it
>>>>>> up
>>>>>> as a language change proposal if you like.
>>>>>>
>>>>>> Ian
>>>>>>
>>>>>
>>>>> > If you actually write `true`, then,
>>>>> > following the usual rules for untyped constant expressions,
>>>>> > it will receive the type `bool`,
>>>>>
>>>>> But, in my expression, the result of a constant expression is still
>>>>> untyped.
>>>>> And "true" is a per-declared untyped boolean value.
>>>>>
>>>>> And, here is another example to should the untyped result is converted
>>>>> to type "bool":
>>>>>
>>>> and sorry again, here "should", I mean "show".
>>>>
>>>>
>>>>>
>>>>> package main
>>>>>
>>>>> type T bool
>>>>>
>>>>> func f() T {return T(false)}
>>>>>
>>>>> func main() {
>>>>> switch true == true {
>>>>> case f(): // invalid case f() in switch (mismatched types T and
>>>>> bool)
>>>>> }
>>>>> }
>>>>>
>>>>>
>>>>> --
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>>>>
>>>
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