On Wed, Nov 18, 2015 at 8:15 AM, Francisco Jerez wrote:
> Matt Turner writes:
>
>> Since the types of the expression were
>>
>>bool ? src_reg : (bool ? brw_reg : brw_reg)
>>
>> the result of the second (nested) ternary would be implicitly
>> converted to a src_reg by the src_reg(struct brw_re
Matt Turner writes:
> Since the types of the expression were
>
>bool ? src_reg : (bool ? brw_reg : brw_reg)
>
> the result of the second (nested) ternary would be implicitly
> converted to a src_reg by the src_reg(struct brw_reg) constructor. I.e.,
>
>bool ? src_reg : src_reg(bool ? brw_r
On Thursday, November 12, 2015 05:46:58 PM Matt Turner wrote:
> Since the types of the expression were
>
>bool ? src_reg : (bool ? brw_reg : brw_reg)
>
> the result of the second (nested) ternary would be implicitly
> converted to a src_reg by the src_reg(struct brw_reg) constructor. I.e.,
>
Since the types of the expression were
bool ? src_reg : (bool ? brw_reg : brw_reg)
the result of the second (nested) ternary would be implicitly
converted to a src_reg by the src_reg(struct brw_reg) constructor. I.e.,
bool ? src_reg : src_reg(bool ? brw_reg : brw_reg)
In the next patch, I
