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Subject: Re: ERROR: INT DW_ATE_unsigned_1 Error emitting BTF type
To:     Mark Wieelard <mark@klomp.org>
CC:     <sedat.dilek@gmail.com>, Masahiro Yamada <masahiroy@kernel.org>,
        Arnaldo Carvalho de Melo <acme@kernel.org>,
        Arnaldo Carvalho de Melo <arnaldo.melo@gmail.com>,
        <dwarves@vger.kernel.org>,
        Linux Kernel Mailing List <linux-kernel@vger.kernel.org>,
        <bpf@vger.kernel.org>, Jiri Olsa <jolsa@kernel.org>,
        Jan Engelhardt <jengelh@inai.de>,
        Domenico Andreoli <cavok@debian.org>,
        Matthias Schwarzott <zzam@gentoo.org>,
        Andrii Nakryiko <andriin@fb.com>,
        Paul Moore <paul@paul-moore.com>,
        Ondrej Mosnacek <omosnace@redhat.com>,
        =?UTF-8?Q?Daniel_P=2e_Berrang=c3=a9?= <berrange@redhat.com>,
        Tom Stellard <tstellar@redhat.com>
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From:   Yonghong Song <yhs@fb.com>
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Date:   Sat, 6 Feb 2021 09:53:31 -0800
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On 2/6/21 8:24 AM, Mark Wieelard wrote:
> Hi,
> 
> On Sat, Feb 06, 2021 at 12:26:44AM -0800, Yonghong Song wrote:
>> With the above vmlinux, the issue appears to be handling
>> DW_ATE_signed_1, DW_ATE_unsigned_{1,24,40}.
>>
>> The following patch should fix the issue:
> 
> That doesn't really make sense to me. Why is the compiler emitting a
> DW_TAG_base_type that needs to be interpreted according to the
> DW_AT_name attribute?
> 
> If the issue is that the size of the base type cannot be expressed in
> bytes then the DWARF spec provides the following option:
> 
>      If the value of an object of the given type does not fully occupy
>      the storage described by a byte size attribute, the base type
>      entry may also have a DW_AT_bit_size and a DW_AT_data_bit_offset
>      attribute, both of whose values are integer constant values (see
>      Section 2.19 on page 55). The bit size attribute describes the
>      actual size in bits used to represent values of the given
>      type. The data bit offset attribute is the offset in bits from the
>      beginning of the containing storage to the beginning of the
>      value. Bits that are part of the offset are padding.  If this
>      attribute is omitted a default data bit offset of zero is assumed.
> 
> Would it be possible to use that encoding of those special types?  If

I agree with you. I do not like comparing me as well. Unfortunately, 
there is no enough information in dwarf to find out actual information.
The following is the dwarf dump with vmlinux (Sedat provided) for
DW_ATE_unsigned_1.

0x000e97e9:   DW_TAG_base_type
                 DW_AT_name      ("DW_ATE_unsigned_1")
                 DW_AT_encoding  (DW_ATE_unsigned)
                 DW_AT_byte_size (0x00)

There is no DW_AT_bit_size and DW_AT_bit_offset for base type.
AFAIK, these two attributes typically appear in struct/union members
together with DW_AT_byte_size.

Maybe compilers (clang in this case) can emit DW_AT_bit_size = 1
and DW_AT_bit_offset = 0/7 (depending on big/little endian) and
this case, we just test and get DW_AT_bit_size and it should work.

But I think BTF does not need this (DW_ATE_unsigned_1) for now.
I checked dwarf dump and it is mostly used for some arith operation
encoded in dump (in this case, e.g., shift by 1 bit)

0x000015cf:   DW_TAG_base_type
                 DW_AT_name      ("DW_ATE_unsigned_1")
                 DW_AT_encoding  (DW_ATE_unsigned)
                 DW_AT_byte_size (0x00)

0x00010ed9:         DW_TAG_formal_parameter
                       DW_AT_location    (DW_OP_lit0, DW_OP_not, 
DW_OP_convert (0x000015cf) "DW_ATE_unsigned_1", DW_OP_convert 
(0x000015d4) "DW_ATE_unsigned_8", DW_OP_stack_value)
                       DW_AT_abstract_origin     (0x00013984 "branch")

Look at clang frontend, only the following types are encoded with 
unsigned dwarf type.

   case BuiltinType::UShort:
   case BuiltinType::UInt:
   case BuiltinType::UInt128:
   case BuiltinType::ULong:
   case BuiltinType::WChar_U:
   case BuiltinType::ULongLong:
     Encoding = llvm::dwarf::DW_ATE_unsigned;
     break;


> not, can we try to come up with some extension that doesn't require
> consumers to match magic names?
> 
> Thanks,
> 
> Mark
>