Consider coercing them to int64, multiply, then shift and coerce back to
Int26_6. I suspect that would be faster than splitting with shifts and
ANDs, two int32 multiplies, shift and add.
On Tue, Jun 14, 2016, 11:42 Dave MacFarlane <driu...@gmail.com> wrote:
> I'm not actually trying to do x & -1, that would be pointless, as you say.
> It was just
> the easiest way to demonstrate the behaviour that I didn't understand in a
> minimal
> way. I understand the problem now--I was thinking of 0x as a prefix
> representing a bitmask
> when used as a constant with a bitwise operation, while Go thinks of it as
> a prefix
> representing a hexadecimal number even in that context.
>
> What I *really* want to do is multiply 2 x/image/math/fixed.Int26_6
> variables. I don't
> want to lose the precision that x*y >> 6 would unnecessarily as x or y get
> large, so I wanted
> to extract the first 26 bits, multiply them, and then separately multiply
> the decimal portion and add
> it back shifted into the correct location.
>
> (Int26_6 is defined as `type Int26_6 int32`)
>
> On Tue, Jun 14, 2016 at 2:16 PM, Ian Lance Taylor <i...@golang.org> wrote:
>
>> 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
>> >>>
>> >>> --
>> >>> You received this message because you are subscribed to the Google
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>> >>
>> >> --
>> >>
>> >> -j
>> >
>> >
>> >
>> >
>> > --
>> > - Dave
>> >
>> > --
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>
>
>
> --
> - Dave
>
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