Why is UInt32 << 32 a no-op?
Why is UInt32 << 32 a no-op?
- Subject: Why is UInt32 << 32 a no-op?
- From: Jens Alfke <email@hidden>
- Date: Tue, 17 Nov 2009 08:56:08 -0800
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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