In the given test case, the loop is split into vectorised and non
vectorised
versions due to peeling. At the end of the loop the IVs are incremented to
their latest value. This is achieved by taking the base of the loop (g_21)
and
adding the iterations (240) multiplied by the step (_6):
:
# _1
On Mon, Aug 1, 2016 at 11:24 AM, Alan Hayward wrote:
> In the given test case, the loop is split into vectorised and non
> vectorised
> versions due to peeling. At the end of the loop the IVs are incremented to
> their latest value. This is achieved by taking the base of the loop (g_21)
> and
> ad
On 01/08/2016 14:49, "Richard Biener" wrote:
>On Mon, Aug 1, 2016 at 11:24 AM, Alan Hayward
>wrote:
>> In the given test case, the loop is split into vectorised and non
>> vectorised
>> versions due to peeling. At the end of the loop the IVs are incremented
>>to
>> their latest value. This is