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Re: [patch] Cut memory address width
On Wed, 27 Sep 2006 20:22:11 +0200, Daniel Jacobowitz wrote:
> On Wed, Sep 27, 2006 at 11:20:22AM -0700, Michael Snyder wrote:
> > On Wed, 2006-09-27 at 18:15 +0200, Jan Kratochvil wrote:
> > > Hi,
> > >
> > > `x/x $ebx' on gdb/amd64 debugging inferior/i386 causes Cannot access memory at
> > > address 0xffffce70 (or so) as $ebx is considered `int' and sign-extended to
> > > 64-bit while the resulting address 0xffffffffffffce70 fails to be accessed.
...
> What's interesting is why this behavior is different on x86_64 and
> i386. Where are we doing the sign extension - that's probably where it
> should be fixed, if anywhere.
In such case `paddress' should print full 64-bit addresses.
Currently it is weird as it refuses to access memory while it will show you
that you were accessing 0xffffce70 - the already cut form - it lies.
If you type on i386 gdb that you want to `x/x 0xffffffffffffce70' it works - as
it will cut the address automatically as even the native C compiler would do.
With these two existing behaviors I was feeling it is more expected to cut the
address even for the access. Still I like more the strict forbidding access
and so I would rather like to remove the existing address-cut in `paddress' and
also wherever it exists for the native i386.
Still the current state is inconsistent, no matter which way to go is the right
one.
Regards,
Jan