Re: [PATCH v5 2/2] Implement pahole-like 'ptype /o' option

[Pedro Alves <palves@redhat.com>]

On 12/13/2017 09:22 PM, Pedro Alves wrote:
> On 12/13/2017 08:36 PM, Sergio Durigan Junior wrote:
>> On Wednesday, December 13 2017, I wrote:
> 
>>> OK, I'll confirm on PPC64BE.
> 
> Thanks.
> 
>>
>> It seems like the algorithm for calculating bitfield offsets is not
>> working correctly on BE machines.  This is not only for "ptype /o", but
>> also for regular print commands.  For example, consider this test:
>>
>>   struct tyu
>>   {
>>     int a1 : 1;
>>
>>     int a2 : 3;
>>
>>     int a3 : 23;
>>
>>     char a4 : 2;
>>
>>     int64_t a5;
>>
>>     int a6 : 5;
>>
>>     int64_t a7 : 3;
>>   };
>>
>>   int
>>   main (int argc, char *argv[])
>>   {
>>     struct tyu e;
>>
>>     e.a1 = e.a2 = e.a3 = e.a4 = e.a6 = e.a7 = -1;
>>
>>     return 0;
>>   }
>>
>> After stopping GDB at the "return 0;" line, here's what we see when we
>> print "e" on x86_64:
>>
>>   (gdb) p e
>>   $1 = {a1 = -1, a2 = -1, a3 = -1, a4 = -1 '\377', a5 = 140737488344880, a6 = -1, a7 = -1}
>>
>> While on PPC64BE:
>>
>>   (gdb) p e
>>   $1 = {a1 = -1, a2 = 3, a3 = 3, a4 = 3 '\003', a5 = 70367536153528, a6 = -1, a7 = -1}
>>
> 
> You didn't initialize e.a5, so even the x86_64 version looks
> wrong at first.  You're seeing stack/register garbage in
> the padding holes.
> 
> You should make that "e" a global to make sure all its
> underlying bytes are clear, including padding.  Or memset it.
> The former is easier.
> 
> a2, a3 and a4 in the PPC64 version do look odd.  Though
> maybe that's something do to with the expression you used.
> 
> Does it make a difference if you initialize all fields
> with separate statements, like:
> 
>  e.a1 = -1;
>  e.a2 = -1;
>  etc.

Actually, I notice now that a4 is plain "char", not
"signed char" and that looks like is the issue.

Plain "char" is unsigned on PPC64.  And then given an
expression like:

  e.a1 = e.a2 = e.a3 = e.a4 ...

e.a4 ends up with an unsigned value (3).  And so
a2 and a3 end up with the same value too (3), and then
a1 ends up "-1" anyway because it's a 1-bit field, i.e.,
no matter what you put there, it ends up being -1,
because all it fits in it is the sign bit.

Thanks,
Pedro Alves