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 <javascript:>>
> 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":
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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