On 02/13/2018 07:50 AM, Peter Maydell wrote:
>> + /* We need two overflow bits at the top. Adding room for that is
>> + a right shift. If the exponent is odd, we can discard the low
>> + bit by multiplying the fraction by 2; that's a left shift.
>> + Combine those and we shift right if the exponent is even. */
>> + a_frac = a.frac;
>> + if (!(a.exp & 1)) {
>> + a_frac >>= 1;
>> + }
>> + a.exp >>= 1;
> Comment says "shift right if the exponent is even", but code
> says "shift right by 1 if exponent is odd, by 2 if exponent is even".
>
The last line is dividing the exponent by 2, not shifting the fraction.
r~
