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.maydell@linaro.org>
Reviewed-by: Aurelien Jarno <aurelien@aurel32.net>
Reviewed-by: Richard Henderson <rth@twiddle.net>
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