Thank you for your comment to the bug 16257.
I am new to LLVM, so not all the aspects of LLVM's /"undef"/ seem clear to me yet.
I read and understood the examples from the LLVM documentation:
However, those examples do not cover all of the possible contexts where
/"undef"/ can appear.
E.g., I can't realize why /"fmul undef, undef"/ may not result in/"undef"/
whereas /"mul undef, undef"/ is always folded to /"undef"/. Seems I am missing
something important about floats here.
in the case of mul undef, undef, the two inputs may be anything. In order to fold the mul to undef you have to prove that the output may be anything (any bit pattern). For example it would be wrong to fold "mul 0, undef" to undef since no matter what value you substitute for "undef" the output of the mul will always be zero. Note that in something like "mul undef, undef", where undef occurs more than once, you are allowed to assume that the two undefs are independent of each other, uncorrelated, i.e. in your reasoning you can insert any bit pattern for the first undef and any other bit pattern for the second undef.
Here is a proof that "mul undef, undef" is undef. Let R represent an arbitrary bit pattern. We have to find two bit patterns X and Y such that "mul X, Y" is equal to R. For example set X to 1 and Y to R. This ends the proof.
Now consider "fmul undef, undef". You could try to apply the same reasoning, and maybe it is OK. Let R represent an arbitrary bit pattern. Set Y = R and set X to be the bit pattern for the floating point number 1.0. Then "fmul X, Y" is equal to R, proving that "fmul undef, undef" can be folded to "undef". But... is this correct? Is it always true that "fmul 1.0, R" always has exactly the same bits as R (this is not the same as the output comparing equal to R using a floating point comparison)? After all, R might be something funky like a signed zero, an infinity, or a NaN. I don't know if it is always true if "fmul 1.0, R" always has the same bits as R, hopefully a floating point expert can tell us.
You could always argue that you could choose "undef" to be a signalling NaN, causing the program to have undefined behaviour at the point of the multiplication, in which case you could do anything, but kindly instead just fold the fmul to undef. I don't much like using snans like this though since they can be made non-signalling by poking the processor AFAIK.
The same applies to "fdiv".
/"fdiv undef, undef"/ is not folded but /"div %X, undef"/ and /"div undef, %X"/
are folded to /"undef"/ (SimplifyFDivInst function in
Here is the argument for "div %X, undef -> undef". The undef value might be zero (div %X, 0), so lets choose it to be zero. That means that the program has undefined behaviour at this point, and so we could fold the div to "exit(1)" or "launch nuclear missile". But instead we only fold it to undef.
You are wrong to say that "div undef, %X" is folded to "undef" by InstructionSimplify, it is folded to zero. It would be wrong to fold to undef, since (eg) if %X is 2 then you can't obtain an arbitrary bit pattern as the output.
Fdiv is harder than div because a floating point division by 0.0 has a defined result, unlike for div.
Moreover, SimplifyFDivInst does not take
into account whether signalling of SNaNs can be switched off or not - it always
folds if one of the operands is /"undef"/.
This is either a mistake, or a decision that in LLVM IR snans are always considered to be signalling.
Another mysterious thing for me is folding of /"mul %X, undef"/.
The result depends on whether %X is odd or even:
* "undef" if %X is odd or equal to "undef";
* 0 otherwise.
InstructionSimplify folds "mul %X, undef" to 0 always, since after all "undef" could be zero, in which case the output is 0. It would be wrong to fold it to undef, as you point out, and it isn't folded to undef.
There is a similar bug 16258 about folding of /"fadd undef, undef"/. /"Add"
/gets folded to /"undef"/ if one or both its operands are /"undef"/.
Should /"fadd"/ behave differently than integer /"add"/ too?
It's the same problem. It is easy to show that "add undef, %X" can produce any desired bit pattern as output, but it is less clear whether that is true for fadd, unless you use the snan argument.