similar to: [LLVMdev] a question about constant fold for fdiv

On Thu, 2007-03-22 at 15:50 -0700, leo han wrote: > Hello, I have a question about the constant folding for fdiv instructions. > For the instruction "fdiv double 0.0, 0.0", the folded result is inf. I > think this should be nan. Can anyone tell me why it is not nan? I think the specification says that it is "undefined" so any value will do. inf is just as undefined

Reid Spencer wrote: > On Thu, 2007-03-22 at 15:50 -0700, leo han wrote: > >> Hello, I have a question about the constant folding for fdiv instructions. >> For the instruction "fdiv double 0.0, 0.0", the folded result is inf. I >> think this should be nan. Can anyone tell me why it is not nan? >> > > I think the specification says that it is

Jeff Cohen wrote: > Reid Spencer wrote: >> On Thu, 2007-03-22 at 15:50 -0700, leo han wrote: >> >>> Hello, I have a question about the constant folding for fdiv instructions. >>> For the instruction "fdiv double 0.0, 0.0", the folded result is inf. I >>> think this should be nan. Can anyone tell me why it is not nan? >>>

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Jeff Cohen wrote: > Jeff Cohen wrote: >> Reid Spencer wrote: >>> On Thu, 2007-03-22 at 15:50 -0700, leo han wrote: >>> >>>> Hello, I have a question about the constant folding for fdiv instructions. >>>> For the instruction "fdiv double 0.0, 0.0", the folded result is inf. I >>>> think this should be nan. Can anyone tell

On 08.08.2013, at 18:25, Chad Rosier <chad.rosier at gmail.com> wrote: > I would like to transform X/Y -> X*1/Y. Specifically, I would like to convert: > > define void @t1a(double %a, double %b, double %d) { > entry: > %div = fdiv fast double %a, %d > %div1 = fdiv fast double %b, %d > %call = tail call i32 @foo(double %div, double %div1) > ret void >

I did few transformation in Instruction *InstCombiner::visitFDiv() in an attempt to remove some divs. I may miss this case. If you need to implement this rule, it is better done in Instcombine than in DAG combine. Doing such xform early expose the redundancy of 1/y, which have positive impact to neighboring code, while DAG combine is bit blind. You should be very careful, reciprocal is very

Hi Chad, This is a great transform to do, but you’re right that it’s only safe under fast-math. This is particularly interesting when the original divisor is a constant so you can materialize the reciprocal at compile-time. You’re right that in either case, this optimization should only kick in when there is more than one divide instruction that will be changed to a mul. I don’t have a strong

I believe we were under the impression that InstCombine, as a canonicalized/optimizer, should not increase code size but only reduce it. Minor aside, but you don't need all of fast-math for the IR, just the "arcp" flag, which allows for replacement of division with reciprocal-multiply. On Aug 8, 2013, at 10:21 AM, Shuxin Yang <shuxin.llvm at gmail.com> wrote: > I remember

Hey guys, I am wondering what the semantics for fdiv is, if the denominator is zero. For sdiv, the language reference specifies that this is undefined behavior. For fdiv, the language reference says nothing (hence, I assume that it is defined somehow). So how is it defined? Does it follow IEEE-754 definition, i.e., +Inf if nominator is > 0, -Inf if nominator is < 0, otherwise NaN?

I remember why I didn't implement this rule in Instcombine. It add one instruction. So, this xform should be driven by a redundancy eliminator if you care code size. On 8/8/13 10:13 AM, Shuxin Yang wrote: > I did few transformation in Instruction *InstCombiner::visitFDiv() in > an attempt to remove some divs. > I may miss this case. If you need to implement this rule, it is >

On Aug 8, 2013, at 9:56 AM, Jim Grosbach <grosbach at apple.com> wrote: > Hi Chad, > > This is a great transform to do, but you’re right that it’s only safe under fast-math. This is particularly interesting when the original divisor is a constant so you can materialize the reciprocal at compile-time. You’re right that in either case, this optimization should only kick in when

Hi, Consider the following test case: int foo(float *A, float *B, float *C, int len, int VSMALL) { for (int i = 0; i < len; i++) if (C[i] > VSMALL) A[i] = B[i] / C[i]; } In this test the div operation is conditional but llvm is generating unconditional div for this case: vector.body: ; preds = %vector.body, %vector.ph %index = phi i64 [

I would like to transform X/Y -> X*1/Y. Specifically, I would like to convert: define void @t1a(double %a, double %b, double %d) { entry: %div = fdiv fast double %a, %d %div1 = fdiv fast double %b, %d %call = tail call i32 @foo(double %div, double %div1) ret void } to: define void @t1b(double %a, double %b, double %d) { entry: %div = fdiv fast double 1.000000e+00, %d %mul = fmul

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Seems incorrect but I forget the IEEE fp rules. What if both x and y are infinity? in that case x/y = NAN but you transformation will yield 0 as the result. On 08/08/2013 09:25 AM, Chad Rosier wrote: > I would like to transform X/Y -> X*1/Y. Specifically, I would like to > convert: > > define void @t1a(double %a, double %b, double %d) { > entry: > %div = fdiv fast

Duncan, > Hi Oleg, > >> >> /This is either a mistake, or a decision that in LLVM IR snans >> are always >> considered to be signalling. / >> Yes, this seems to be an agreement to treat "undef" as a SNaN for >> "fdiv". > > "undef" is whatever bit pattern you want it to be, i.e. the compiler > can assume it is any

Hi Duncan, Thank you a lot for your time to provide that great and informative explanation. Now the "undef" logic makes much more sense for me. >> /You are wrong to say that "div undef, %X" is folded to "undef" by InstructionSimplify, it is folded to zero./ My mistake. I meant to say "*f****div* undef, %X" is folded to "undef" (not

Jeff Cohen wrote: > > The standard apparently doesn't explicitly handle 0/0, but the > position of the IEEE appears to be that it should yield a NaN of the > appropriate sign. See > http://standards.ieee.org/reading/ieee/interp/754-1985.html OK, it does explicitly handle it. My reading comprehension seems to be lacking right now :(

On Thu, Aug 8, 2013 at 1:56 PM, Mark Lacey <mark.lacey at apple.com> wrote: > > On Aug 8, 2013, at 9:56 AM, Jim Grosbach <grosbach at apple.com> wrote: > > Hi Chad, > > This is a great transform to do, but you’re right that it’s only safe > under fast-math. This is particularly interesting when the original divisor > is a constant so you can materialize the