a question

["Marius Nita" <shrimpx@gmail.com>]

Hi, I am trying to understand some endian-specific code in GDB (well,
BFD for the purpose of this message). I would greatly appreciate any
help.

I have a rather specific question regarding the code below, which is
in bfd/aout-arm.c line 420. This code seems to deal with bit-level
endianness. E.g. a "char" address holding a bit pattern 00010000
represents 0x10 on one endianness and 0x08 on the other. The r_length
left-shift below shifts data by 5 bits on big-endian and 1 bit on
little-endian. The r_length data will end up occupying 2 bits in the
natptr->r_type[0] byte. On big-endian, they'll be bits 6 and 7, and on
little-endian, they are bits 2 and 3.

My question is: why aren't the bits in r_length "reversed" to conform
with bit-level endianness? For example, if the r_length bits are "10",
this left-shift results in "0100 0000" on big-endian and "0000 0100"
on little-endian. These bit strings are clearly not the reverse of
each other.

Any help would be much appreciated!

Thank you!

-m

  if (bfd_header_big_endian (abfd))
    {
      natptr->r_index[0] = r_index >> 16;
      natptr->r_index[1] = r_index >> 8;
      natptr->r_index[2] = r_index;
      natptr->r_type[0] =
    (  (r_extern ?   RELOC_STD_BITS_EXTERN_BIG: 0)
     | (r_pcrel  ?   RELOC_STD_BITS_PCREL_BIG: 0)
     | (r_neg    ?   RELOC_ARM_BITS_NEG_BIG: 0)
     | (r_length <<  RELOC_STD_BITS_LENGTH_SH_BIG));
    }
  else
    {
      natptr->r_index[2] = r_index >> 16;
      natptr->r_index[1] = r_index >> 8;
      natptr->r_index[0] = r_index;
      natptr->r_type[0] =
    (  (r_extern ?   RELOC_STD_BITS_EXTERN_LITTLE: 0)
     | (r_pcrel  ?   RELOC_STD_BITS_PCREL_LITTLE: 0)
     | (r_neg    ?   RELOC_ARM_BITS_NEG_LITTLE: 0)
     | (r_length <<  RELOC_STD_BITS_LENGTH_SH_LITTLE));
    }