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: Jocelyn Houle <email@hidden>
- Date: Tue, 17 Nov 2009 12:19:21 -0500
The Intel instruction uses a 5-bit value to define the amount of shift, as spec'd here:
http://www.intel.com/Assets/PDF/manual/253667.pdf
"The destination operand can be a register or a memory location. The count operand can be an immediate value or the CL register. The count is masked to 5 bits (or 6 bits if in 64-bit mode and REX.W is used). The count range is limited to 0 to 31 (or 63 if 64-bit mode and REX.W is used). A special opcode encoding is provided for a count of 1."
So, it's a hardware limitation... (an extra sticky bit could have handled all possible cases, I guess).
For your case, you can work around the problem in a number of ways.
Here is the code I've been using to generate bitmask of 0 to 32 bits, inclusive:
inline uint32_t bitmask( uint32_t s )
{
uint32_t s_2 = (s >> 1);
return (1<<s_2<<(s-s_2)) - 1;
}
The code is an extension of a simpler:
mask = (1 << s) - 1;
which works for 0 to 31 bits.
All I did was do 2 successive shifts of half the number of bits. Keeps the code branchless. Some might argue that using a LUT could be faster in certain circumstances, but I never needed to optimize such a code that much.
On 2009-11-17, at 11:56, 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 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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