Re: [help]failed to generate PHI NODE in esra pass.

[Handong Ye <yehandong at gmail dot com>]

On Mon, Nov 5, 2012 at 5:32 AM, Martin Jambor <mjambor@suse.cz> wrote:
> Hi,
>
> On Sun, Nov 04, 2012 at 09:32:48PM -0800, Handong Ye wrote:
>> On Sun, Nov 4, 2012 at 2:13 PM, Martin Jambor <mjambor@suse.cz> wrote:
>> > On Sat, Nov 03, 2012 at 09:01:53AM +0000, Yangyueming wrote:
>> >> Hi, all
>> >>
>
> ...
>
>> >>
>> >> But when I do the test for a case with a little change, it is failed to generate MAX_EXPR in phiopt1.
>> >> The failed case2 :
>> >> int foo(short a ,short b)
>> >> {
>> >>   int c;
>> >>   if (a < b)
>> >>     a = b;
>> >>
>> >>   c = *(int *)&a;
>> >
>> > ehm, and just what do you expect the result of the above line to be?
>> >
>> >>   return c;
>> >> }
>> >>
>
> ...
>
>> >
>> > Nevertheless, I can tell why there is no phi node for a. a has its
>> > address taken and therefore it is not in SSA form (as clearly visible
>> > in the first statements in bb 2 and bb 3).  Thus, no PHI nodes for a.
>> >
>>
>> Why 'a' has no SSA form if it's address taken?
>> I didn't check the code, but I guess it may because 'a' value is not
>> directly used.
>
> Because address taken variables might be used and defined indirectly
> through a memory load/store, often in ways that are practically
> impossible to detect.

Isn't memory ssa supposed to handle such issues? ...
Seems I need do some research ...

Thanks
Handong

>
> Martin
>
>
>>
>> >> Dump of phifail.c.027t.esra:
>> >> foo (short int a, short int b)
>> >> {
>> >>   int c;
>> >>   short int a$0;
>> >>
>> >> <bb 2>:
>> >>   a$0_1 = a;
>> >>   if (a$0_1 < b_2(D))
>> >>     goto <bb 3>;
>> >>   else
>> >>     goto <bb 4>;
>> >>
>> >> <bb 3>:
>> >>   a = b_2(D);
>> >>
>> >> <bb 4>:
>> >>   c_4 = MEM[(int *)&a];
>> >>   c_5 = c_4;
>> >>   return c_5;
>> >>
>> >> }
>> >>
>> >> Why it is failed and if there's a way to make it work?
>> >
>> > I believe that the modified input has undefined semantics (sorry, I
>> > don't have time to verify this in the standard) and so no, there is no
>> > way to make it work.
>> >
>> > Martin
>>
>>
>>
>> --
>> Thanks.
>> Handong



-- 
Thanks.
Handong