On 12/19/2013 01:57 PM, Peter Maydell wrote: > If the input to float*_scalbn() is denormal then it represents > a number 0.[mantissabits] * 2^(1-exponentbias) (and the actual > exponent field is all zeroes). This means that when we convert > it to our unpacked encoding the unpacked exponent must be one > greater than for a normal number, which represents > 1.[mantissabits] * 2^(e-exponentbias) for an exponent field e. > > This meant we were giving answers too small by a factor of 2 for > all denormal inputs. > > Note that the float-to-int routines also have this behaviour > of not adjusting the exponent for denormals; however there it is > harmless because denormals will all convert to integer zero anyway. > > Signed-off-by: Peter Maydell <peter.mayd...@linaro.org> > ---
At least the float128 routine needs the same fix. I'm less certain about the floatx80 routine, since IIRC that format has no implicit 1 bit -- it's always explicit. r~
