Andres Freund <and...@anarazel.de> writes:
> On 2017-04-09 19:20:27 -0400, Tom Lane wrote:
>> As I read that, it's only "undefined" if overflow would occur (ie
>> the sign bit would change). Your compiler is being a useless annoying
>> nanny, but that seems to be the in thing for compiler authors these
>> days.
> "The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are
> filled with
> zeros. If E1 has an unsigned type, the value of the result is E1 × 2 E2 ,
> reduced modulo
> one more than the maximum value representable in the result type. If E1 has a
> signed
> type and nonnegative value, and E1 × 2 E2 is representable in the result
> type, then that is
> the resulting value; otherwise, the behavior is undefined."
> As I read this it's defined iff E1 is signed, nonnegative *and* the the
> result of the shift is representable in the relevant type. That seems,
> uh, a bit restrictive, but that seems to be the only reading?
Oh --- I misread the "nonnegative" as applying to the shift count, but
you're right, it's talking about the LHS. That's weird --- the E1 × 2^E2
definition works fine as long as there's no overflow, so why didn't they
define it like that? It seems just arbitrarily broken this way.
regards, tom lane
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