Re: [PATCH] memattr bounds

[Andrew Cagney <ac131313@cygnus.com>]

> The following addresses edge conditions in the mem command by making a
> special case for upper bound == 0.  When the upper bound is zero, it is
> assumed that the user wants an upper bound of max CORE_ADDR+1. Currently,
> it's not possible to define a memory region with zero as it's upper bound, 
> so this should not conflict with any current usage.

Don't forget to follow up with something for the doco.

> The patch also corrects a bug that allowes the definition of overlapping
> memory regions, where the new region starts below an existing region and
> extends above it. (Or does someone think that is a feature?)

I would treat it as a bug.

Some of the simulators have a mechanism that is similar to memattr where 
the address map is layered and each address range is assigned to a 
layer.  Within a layer things can't overlap but between layers they can 
- this making the search well defined.   Adding such a mechanism to 
memattr would be a useful but perhaps best left to the reader :-)

> 2002-06-21  Don Howard  <dhoward@redhat.com>
> 
> 	* memattr.c (create_mem_region): Treat hi == 0 as a special case
> 	that means max CORE_ADDR+1.
> 	(lookup_mem_region): Ditto.
> 	(mem_info_command): Ditto.
> 
> 


Yes,

1.5 tweaks:

> -      if ((lo >= n->lo && lo < n->hi) ||
> -	  (hi > n->lo && hi <= n->hi))
> +      if ((lo >= n->lo && (lo < n->hi || n->hi == 0)) ||
> +	  (hi > n->lo && (hi <= n->hi || n->hi == 0)) ||
> +	  (lo <= n->lo && (hi >= n->hi || hi == 0)))

The ``||'' goes at the start of the line (the old code was also wrong :-)

>        if (TARGET_ADDR_BIT <= 32)
> -	tmp = local_hex_string_custom ((unsigned long) m->hi, "08l");
> +	tmp = local_hex_string_custom ((unsigned long) (m->hi ? m->hi : ~0), "08l");
>        else
> -	tmp = local_hex_string_custom ((unsigned long) m->hi, "016l");
> +	tmp = local_hex_string_custom ((unsigned long) (m->hi ? m->hi : ~0), "016l");
>        

Consider lifting the ``(m->hi ? m->hi : ~0)'' out to before the if().

which ever, it can then go straight in.
Andrew