On 20 December 2013 17:34, Richard Henderson wrote:
> 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
>
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
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 represen
