On 13 February 2018 at 16:23, Richard Henderson
<richard.hender...@linaro.org> wrote:
> 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.
Doh, so it is.
-- PMM
