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Re: bfd question writes:

> It seems that most uses of this struct in the code are guarded by a
>  bfd_header_big_endian(...)
> check. E.g. line 420 in bfd/aout-arm.c. If bfd_header_big_endian() is
> true, it's assumed that the three bytes in r_index are laid out
> right-to-left. When it's false, they are left-to-right. Also,
> depending on the value of bfd_header_big_endian(), the *bits* in
> r_type[0] are laid out differently -- right to left vs left to right.
> Could someone help me understand why this is the case?

Don't try to understand a.out, it is mostly defined by legacy. :-)

> Is it fundamentally tied to the endianness of binary formats? If so,
> why are the *bits* inside r_type[0] reversed?

r_type is actually a bitfield, and bitfield endianess on most targets
coincides with byte endianess.


Andreas Schwab, SuSE Labs,
SuSE Linux Products GmbH, Maxfeldstraße 5, 90409 Nürnberg, Germany
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