[LLVMdev] Shifts that use only 5 LSBs.

[Daniel M Gessel]

I can't find the bug you refer to.

Also, it doesn't have this problem in x86: it uses the shldl  
instruction.

PPC32, interestingly enough, generates something similar, but looks  
like it has extra instructions to or in what's guaranteed to be 0.

Reminding myself of some PPC assembler though, so I'm not 100%.

Thanks,

Dan




On Dec 16, 2008, at 9:27 PM, Eli Friedman wrote:

> On Tue, Dec 16, 2008 at 5:20 PM, Daniel M Gessel <gessel at apple.com>  
> wrote:
>> The problem here is that it looks like LLVM is introducing an  
>> expansion that
>> assumes 32 bit shifts use more than 5 bits of the shift value.
>> I created a simple test function:
>> u64 mebbe_shift( u64 x, int test )
>> {
>> if( test )
>> x <<= 2;
>> return x;
>> }
>> I compile using clang, opt, and llc.
>> I get something that, converted from my assembler to hasty psuedo-C:
>> u64 mebbe_shift( u64 x, int test )
>> {
>> int amt = test ? 2 : 0;
>> x.hi = x.hi << amt  |  x.lo >> (32 - amt);
>> x.lo <<= amt;
>>
>> return x;
>> }
>
> Ouch, that's nasty... I just filed http://llvm.org/bugs/show_bug.cgi?id=3225.
>
> -Eli
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