Re: Can this be happening?

Jim Blandy
2007-07-03T21:38:48Z

"Mohammed, Moqtadir" <Moqtadir_Mohammed@reyrey.com> writes:
> I was trying to look at a core dump of a program, and gdb displays the following result for
> #info registers
>
> eax 0xa0 160
> ecx 0x2 2
> edx 0xa 10
> ebx 0xa7e3de9c -1478238564
> esp 0xa6babddc 0xa6babddc
> ebp 0xa6babe00 0xa6babe00
> esi 0xa7ef9d9a -1477468774
> edi 0x838f44c 137950284
> eip 0xa7d85cec 0xa7d85cec <mempcpy+28>
> eflags 0x50203 [ CF IF RF AC ]
> cs 0x73 115
> ss 0x7b 123
> ds 0x7b 123
> es 0xb010007b -1341128581
> fs 0x0 0
> gs 0x33 51
>
> Platform: IA32. (elf)
>
> My question is, how is the register 'es' being reported as a 32 bit value.
> I may be completely naive asking this question, but I have been trying to google for anything
> related to it, but haven't found an answer. Is 'es' not supposed to be only 16bit.

Well, actually, GDB thinks they're all 32 bits long:

static struct type *
i386_register_type (struct gdbarch *gdbarch, int regnum)
{
if (regnum == I386_EIP_REGNUM)
return builtin_type_void_func_ptr;

if (regnum == I386_EFLAGS_REGNUM)
return i386_eflags_type;

if (regnum == I386_EBP_REGNUM || regnum == I386_ESP_REGNUM)
return builtin_type_void_data_ptr;

if (i386_fp_regnum_p (regnum))
return builtin_type_i387_ext;

if (i386_mmx_regnum_p (gdbarch, regnum))
return i386_mmx_type (gdbarch);

if (i386_sse_regnum_p (gdbarch, regnum))
return i386_sse_type (gdbarch);

#define I387_ST0_REGNUM I386_ST0_REGNUM
#define I387_NUM_XMM_REGS (gdbarch_tdep (current_gdbarch)->num_xmm_regs)

if (regnum == I387_MXCSR_REGNUM)
return i386_mxcsr_type;

#undef I387_ST0_REGNUM
#undef I387_NUM_XMM_REGS

return builtin_type_int;
}

You can always check in GDB:

(gdb) p sizeof ($cs)
$1 = 4
(gdb) ptype $cs
type = int
(gdb)

I don't know why the upper bits would be set. GDB may be
misinterpreting the information in the core file.