[Dwarf-Discuss] DWARF on systems where memory is not byte addressable

Chris Quenelle chris.quenelle at oracle.com
Thu Jul 26 11:54:47 PDT 2012


I don't really work in this area anymore, but I can wave my
hands around with the best of them, so here's my two cents.

It sounds like you are asking this question:

Is it easier to teach gdb that a byte is 32-bits long, or is it
easier to teach gdb that pointers are always missing their bottom
two bits (which would always be zero anyway).
(I would suggest taking one of those two approaches consistently,
otherwise it will be very hard to figure out what's going on.)

If that's really what your question boils down to, then I would look
first to the gdb code and decide which approach is better in gdb,
and then decide what possible tweaks you might need to make
in gcc dwarf generation to support those changes.

It doesn't seem like a formal definition of "what a byte is"
is going to actually help make any specific software work.
So that might be a red herring.

--chris




On Wednesday July 25, at    10:45PM, Joeri van Ruth wrote:

> Hello all, I am wondering about how to deal with platforms with word
> memories, by which I mean that the smallest addressable unit in memory
> is (in our current case) 32 bits wide.  This means that at the C level, 
> 
> 	sizeof(char) == sizeof(short) == sizeof(int) == 1,
> 
> so far so good.  However, we are having problems with gdb.  I am aware
> that this may be entirely gdb specific but I do note that the standard
> does not spend a lot of words on the issues that arise here, that's
> why I bring this up here.
> 
> The standard does not seem to define anywhere how large a byte is
> supposed to be.  Historically, older architectures used anything out
> of 6, 7, 8 and 9 bits bytes which is why networking standards tend to
> speak of octets instead.  DWARF seems to assume 8 bit bytes, hence the
> LEB128 encoding, but it does not state so explicitly unless I
> overlooked something.
> 
> A C oriented view might consider that sizeof(char) == sizeof(int), and
> as C does not distinguish clearly between byte and char, take a byte
> to be 32 bits wide.  But even that's not always the case as sometimes
> we see word oriented platforms which still take the arithmetic size of
> char to be 8 bits, requiring frequent sign- and zero-extension when
> assigning to a char or short variable.
> 
> However, I assume that if the DWARF standard were explicit about the
> size of a byte, it would define a byte to be 8 bits.
> 
> The problem we see with gdb hinges on the DW_AT_byte_size attribute of
> a type descriptor.  Gdb uses it for at least two purposes:
> 
> 	- to perform address arithmetic
> 
> 	- to determine the bit size of values
> 
> If we set the DW_AT_byte_size of an integer to 1, gdb will do the
> address arithmetic correctly, that is, look for int_array[1] at
> address int_array + 1, not + 4, but if you ask for the value of an int
> variable it will only display the lower 8 bits.
> 
> If we set the DW_AT_byte_size of int to 4, which indeed sounds
> more consistent given the name _byte_size, gdb will extract the full
> 32 bits of the value but get the address arithmetic wrong as
> int_array[1] now actually accesses int_array[4].
> 
> It seems to me that the proper way would be to fix gdb to take the
> addressing size unit into account as general knowledge of the target
> platform, but I can't believe we're the first to come across this.  I
> wonder if anyone on this list has already faced similar issues and
> what they did about it.
> 
> Best regards,
> 
> Joeri van Ruth
> 
> -- 
> Joeri van Ruth, ACE Associated Compiler Experts
> De Ruyterkade 113, 1011 AB Amsterdam, The Netherlands.
> Tel: +31 20 6646416, Fax: +31 20 6750389,
> mailto:joeri at ace.nl, http://www.ace.nl
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