The current code appears to work in simple tests, however this will
guarantee that the returned exponent is 0 for a 0 value.
Signed-off-by: Ilia Mirkin
---
I couldn't make a simple test-case that would cause the current logic to
fail. However when I did the same thing with doubles, I ran into tr
On Fri, Jul 18, 2014 at 9:19 AM, Ilia Mirkin wrote:
> The current code appears to work in simple tests, however this will
> guarantee that the returned exponent is 0 for a 0 value.
>
> Signed-off-by: Ilia Mirkin
> ---
>
> I couldn't make a simple test-case that would cause the current logic to
>
On Fri, Jul 18, 2014 at 5:27 PM, Matt Turner wrote:
> On Fri, Jul 18, 2014 at 9:19 AM, Ilia Mirkin wrote:
>> The current code appears to work in simple tests, however this will
>> guarantee that the returned exponent is 0 for a 0 value.
>>
>> Signed-off-by: Ilia Mirkin
>> ---
>>
>> I couldn't ma
On Fri, Jul 18, 2014 at 2:37 PM, Ilia Mirkin wrote:
> On Fri, Jul 18, 2014 at 5:27 PM, Matt Turner wrote:
>> For 0.0f, applying the f2i and abs out of order doesn't affect the
>> result, but for -0.0f (0x8000, -2147483648) instead of getting 0,
>> you'd get abs(-2147483648) (which is likely -
