how to resolve segment:offset adres in GDB

Question

(gdb) info registers ds
   ds             0x7b  123
(gdb) disassemble
   Dump of assembler code for function printf@plt:
   0x0804831c <+0>: jmp    DWORD PTR ds:0x804a008
=> 0x08048322 <+6>: push   0x10
   0x08048327 <+11>:    jmp    0x80482ec
End of assembler dump.

Can someone describe me how to map ds:0x804a008 address into linear address I can use i "x/xw address" command? If it is not clear I'd like to know where to this first jmp function in code jumps.

Answer 1

Modern x86 OS don't use segmented addressing. Real mode segmented address can only represent 1Mb of address space. This addressing scheme is only used during the boot process for compatibility reasons.

The OS set all the segment registers to a selector that represent the flat 32-bit address space of your process but you shouldn't have to worry about that.

ds:0x804a008 is just the same as 0x804a008

Answer 2

0x804a008 is an address in the processes linear address space - the DWORD in that memory location is the address that will be jumped to (ie., 0x804a008 is a pointer). `

So

x/xw 0x804a008

will dump the contents of the pointer, and

disasm *0x804a008

will disassemble the code that jumping through that pointer will execute.