On 7/6/20 4:47 PM, Max Filippov wrote: > pickNaNMulAdd logic on Xtensa is the same as pickNaN when applied to > the expression (a * b) + c. So with two pickNaN variants there must be > two pickNaNMulAdd variants.
"Is the same as"? I question the non-use of the infzero parameter. When infzero, (a * b) = (Inf * 0), which will produce a default QNaN. Your sentence above would suggest that pickNaN is applied twice, so that if use_first_nan, the default nan is chosen above any nan within c. In addition, is the invalid flag raised for (Inf * 0) + NaN? Does that happen regardless of the use_first_nan setting, or does the whole operation short-circuit? r~
