On Tue, Jun 14, 2016 at 10:44 AM, Dave MacFarlane <driu...@gmail.com> wrote:
> What I'm not understanding is why that's not the case in this example. It's
> a 32 bit bitwise operation on a 32 bit signed type. Shouldn't 0xFFFFFFFF be
> coerced to a value of -1?
Why don't you just write -1?
I don't actually understand what you are doing. Given an int32 value
x, x & 0xFFFFFFF (assuming that were valid) is always simply x. What
else could it be? If you want to mask out the sign bit you should
write x & 0x7FFFFFFF.
Ian
> On Tue, Jun 14, 2016 at 1:23 PM, Jan Mercl <0xj...@gmail.com> wrote:
>>
>> 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
>>>
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>>
>> --
>>
>> -j
>
>
>
>
> --
> - Dave
>
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