[Bug gdb/19487] New: Result of call to __errno_location() is truncated to 32 bit on amd64

["b7.10110111 at gmail dot com" <sourceware-bugzilla at sourceware dot org>]

https://sourceware.org/bugzilla/show_bug.cgi?id=19487

            Bug ID: 19487
           Summary: Result of call to __errno_location() is truncated to
                    32 bit on amd64
           Product: gdb
           Version: 7.10
            Status: NEW
          Severity: normal
          Priority: P2
         Component: gdb
          Assignee: unassigned at sourceware dot org
          Reporter: b7.10110111 at gmail dot com
  Target Milestone: ---

Consider the following C program:

#include <stdio.h>
#include <errno.h>
int main()
{
    printf("%p\n",&errno);
}

If I run it in gdb, I get wrong results of attempt to print errno and even just
its location. Here's the debugging session log:

-------------------------------------------------
$ gdb -q ./gdb-bug -ex 'b main' -ex r
Reading symbols from ./gdb-bug...(no debugging symbols found)...done.
Breakpoint 1 at 0x40059a
Starting program: /tmp/gdb-bug 

Breakpoint 1, 0x000000000040059a in main ()
(gdb) p/x __errno_location()
$1 = 0xf7fbf6a0
(gdb) p *__errno_location()
Cannot access memory at address 0xfffffffff7fbf6a0
(gdb) c
Continuing.
0x7ffff7fbf6a0
[Inferior 1 (process 869) exited normally]
----------------------------------------------------

See that "p/x" truncates the value to 32 bits, then "p *..." sign-extends the
result of truncation, while on "c" we get actual value printed by the program
itself, which is neither of the former two.

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