[llvm-dev] Need help in understanding llvm optimization

[Craig Topper via llvm-dev]

The compiler constant folded it as an optimization. Was this compiled with
optimization’s enabled?

On Sat, Aug 11, 2018 at 1:15 PM SANGEETA CHOWDHARY <
sc1696 at scarletmail.rutgers.edu> wrote:

> I am sorry to not make my question clear. My question is how this value is
> calculated without having fadd and fsub in IR?
>
> On Sat, Aug 11, 2018 at 4:06 PM, Craig Topper <craig.topper at gmail.com>
> wrote:
>
>> Neither 1e16 or 1e16+1 can be accurately represented in a double. The
>> largest integer than be accurately represented is 2^53. As the number gets
>> larger floating sacrifices precision in the lower digits. Because of this
>> 1e16 and 1e16+1 end up having the same representation. So the result of
>> subtracting them is 0.0.
>>
>> On Sat, Aug 11, 2018 at 12:30 PM SANGEETA CHOWDHARY via llvm-dev <
>> llvm-dev at lists.llvm.org> wrote:
>>
>>> Hi,
>>>
>>> I have below code in C -
>>>
>>> int main() {
>>>
>>>   double x,y;
>>>
>>>   x = 1e16;
>>>
>>>   y = (x + 1) - x;
>>>
>>>   printf("y:%e\n", y);
>>>
>>>   return 0;
>>>
>>> }
>>>
>>> llvm bitcode looks like this for this function -
>>>
>>>
>>> ; Function Attrs: nounwind uwtable
>>>
>>> define dso_local i32 @main() local_unnamed_addr #0 {
>>>
>>> entry:
>>>
>>>   %call = tail call i32 (i8*, ...) @printf(i8* getelementptr inbounds ([
>>> 4 x i8], [4 x i8]* @.str, i64 0, i64 0), double 0.000000e+00)
>>>
>>>   ret i32 0
>>>
>>> }
>>>
>>> I am not able to understand how addition and subtraction are performed
>>> in this code. There is no fadd or fsub instruction. How llvm knows that
>>> result of y is 0?
>>>
>>> Is there any way to disable this in llvm?
>>>
>>>
>>> Any help would be much appreciated.
>>>
>>>
>>> Regards,
>>>
>>> Sangeeta
>>> _______________________________________________
>>> LLVM Developers mailing list
>>> llvm-dev at lists.llvm.org
>>> http://lists.llvm.org/cgi-bin/mailman/listinfo/llvm-dev
>>>
>> --
>> ~Craig
>>
>
> --
~Craig
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