Re: Why is UInt32 << 32 a no-op?
Re: Why is UInt32 << 32 a no-op?
- Subject: Re: Why is UInt32 << 32 a no-op?
- From: "Eric A. Borisch" <email@hidden>
- Date: Tue, 17 Nov 2009 12:49:30 -0600
If you're on a 64-bit platform, this shouldn't have much (if any) of a
performance hit:
n = (UInt64) n << shift
On Tue, Nov 17, 2009 at 10:56 AM, Jens Alfke <email@hidden> wrote:
> Surprisingly, the expression "n << 32", where n is a 32-bit integer, is a no-op. I would expect it to result in zero for all values of n.
>
> UInt32 n = (UInt32)-1L;
> int shift = 32;
> n = n << shift;
> assert(n==0); // fails; n is actually unchanged
>
> This is breaking some code of mine that masks out the upper b bits of a number, where 0≤b≤8; it fails when b==0 because the mask it generates is all 1s instead of all 0s. I'm going to have to add a special case.
>
> Any C expert know whether this is in-spec or not?
>
> —Jens _______________________________________________
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--
Eric A. Borisch
email@hidden
Howard Roark laughed.
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