64bit pointer

[Andreas Schwab]

freindlyuser@hushmail.com writes:

> I didn't know if this was a bug or if I was just doing something
> wrong. Basically when I try to use * to get the value pointed to by
> something it gives me 32 bits of the pointer which is actually a
> 64bit address.
>
>
> (gdb) x/s *($rdi+0x8)
> 0x4210a9b7:      <Address 0x4210a9b7 out of bounds>
> (gdb) x/x $rdi+0x8
> 0x3a9b4210b7a4: 0x00003a9b4210a9b7
> (gdb) x/s 0x00003a9b4210a9b7
> 0x3a9b4210a9b7:  "The string it points to"
>
> In the above I wanted to read the string pointed to by the pointer
> at $rdi+0x8
>
> Should this be happening?

Since the type of $rdi is not a pointer, gdb is just being helpful and
implicitly converts the value to (int *) before applying the indirection
operator.

> Should I be using a different command or something different to the
> asterisk?
> Is there a work around where I can read the data in
> 0x00003a9b4210a9b7 without having to manually copy and paste (ie:
> in the commands that are executed on a breakpoint).

There are several options:

- Add a cast, {type} is short for *(type *):

(gdb) x/s {char *}($rdi+8)

- Use the $__ convenience variable after examining the address:

(gdb) x/x $rdi+8
(gdb) x/s $__

Andreas.

-- 
Andreas Schwab, schwab@redhat.com
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