search for: fpbinop

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2020 Feb 07
3
Why does FPBinOp(X, undef) -> NaN?
I came across this comment in SelectionDAG.cpp: case ISD::FADD: case ISD::FSUB: case ISD::FMUL: case ISD::FDIV: case ISD::FREM: // If both operands are undef, the result is undef. If 1 operand is undef, // the result is NaN. This should match the behavior of the IR optimizer. That isn't intuitive to me. I would have expected a binary FP operation with one undef operand to
2020 Feb 07
2
Why does FPBinOp(X, undef) -> NaN?
On Fri, Feb 7, 2020 at 12:29 PM Nuno Lopes <nunoplopes at sapo.pt> wrote: > > It's not correct (output of Alive2): > > define half @fn(half %a) { > %b = fadd half %a, undef > ret half %b > } > => > define half @fn(half %a) { > ret half undef > } > Transformation doesn't verify! > ERROR: Value mismatch > > Example: > half %a