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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