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Re: [PATCH 4/5]: Enhancements to "flags": i386 cleanup


On 10/06/2016 02:44 PM, Anton Kolesov wrote:
> Hi Pedro,
> 
>>
>> Note how that left several flags with 2-bit and/or 4-bit
>> long bitfields:
>>
>>    tdesc_add_bitfield (type, "E", 1, 2);
>>    tdesc_add_bitfield (type, "A", 3, 4);
>> ...
>>    tdesc_add_bitfield (type, "E", 1, 4);
>>
>> which I understand means these two fields will
>> be given uint32_t type instead of bool?  What does this
>> mean in practice?  E.g,. for "A", what do we print when both
>> bits 3 and 4 are clear?  What do we print if one
>> of the bits is set and the other is clear?
> 
> With regards of ARC flags, if field is longer than one bit, then it should be
> treated as an uint. For example, in arc-v2.c field H means "halt bit", so
> it is a single bit, but E is a "Interrupt priority level", so bits are not
> independent in this field - it is a 4-bit integer number, there is no idea
> of independent "first bit" or "second bit" inside this field. If there would,
> then I'd split it into separate fields bits. So it should be printed something
> like "[ H E=1 AE ]" - bits printed only when they are set, uint fields are
> printed as "name=value", though I'm not sure if it should be printed if value
> is 0.  At least that is what are my expectations of how "flags" register should
> be presented.

Thanks.  Looks like EL on aarch64 is similar.  It's an exception
level, I believe.

To confirm what happens with uint bitfields within flags, I hacked
my local x86-64 GDB with:

--- c/gdb/features/i386/64bit-core.xml
+++ w/gdb/features/i386/64bit-core.xml
@@ -10,8 +10,7 @@
   <flags id="i386_eflags" size="4">
     <field name="CF" start="0" end="0"/>
     <field name="" start="1" end="1"/>
-    <field name="PF" start="2" end="2"/>
-    <field name="AF" start="4" end="4"/>
+    <field name="PF" start="2" end="4"/>
     <field name="ZF" start="6" end="6"/>
     <field name="SF" start="7" end="7"/>
     <field name="TF" start="8" end="8"/>

and (after regenerating the gdb/feature/ .c files.), I see:

(gdb) p $eflags = 0
$1 = [ PF=0 ]
(gdb) p $eflags = 0xffffffff
$2 = [ CF PF=7 ZF SF TF IF DF OF NT RF VM AC VIF VIP ID ]

So =0 is always shown for these.  Debatable, but that seems
like just a presentation thing.  Sorry for all my confusions.
I'll go close the PR, and unblock 7.12!

Thanks,
Pedro Alves


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