All binary operators, except shifts, require identical left and right
types. Untyped values will be coerced to the type of the other side, if
representable as such after the conversion. That's not the case in this
example.
On Tue, Jun 14, 2016, 19:16 Dave MacFarlane <driu...@gmail.com> wrote:
> Is this supposed to be legal in Go:
>
> var x int32 = 3
>
> fmt.Printf("%d", x & 0xFFFFFFFF)?
>
> The language spec just says the bitwise operator "applies to integers
> only" and
> "yields a result of the same type as the first operand" that I can see,
> but it's giving
> me a compiler error:
>
> ./main.go:10: constant 4294967295 overflows int32
>
> with go 1.6.2.
>
> Is this a compiler bug, or am I missing something else in the spec that
> makes it impossible
> to mask out the high bit in a signed integer type without converting to an
> unsigned equivalent first?
>
> - Dave
>
> --
> You received this message because you are subscribed to the Google Groups
> "golang-nuts" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> email to golang-nuts+unsubscr...@googlegroups.com.
> For more options, visit https://groups.google.com/d/optout.
>
--
-j
--
You received this message because you are subscribed to the Google Groups
"golang-nuts" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to golang-nuts+unsubscr...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.
