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: email@hidden (Christian Demmer)
- Date: Tue, 17 Nov 2009 18:17:36 +0100
Jens Alfke 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 0â¤bâ¤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?
Shift is undefined if the right operand is not less than the number of
bits in n. So the behavior is in spec.
It's also reasonable; the compiler is probably doing (shift %
32) to calculate the shift value.
Greetings, Christian
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