I need to understand the add instruction in assembly code:
=> 0x08048bff <+43>: add 0x14(%esp,%ebx,4),%eax
(gdb) i r
eax 0x1 1
ecx 0x0 0
edx 0x0 0
ebx 0x1 1
esp 0xffffcd70 0xffffcd70
ebp 0xffffcdc8 0xffffcdc8
esi 0x0 0
edi 0x0 0
eip 0x8048bff 0x8048bff <phase_2+43>
eflags 0x202 [ IF ]
cs 0x23 35
ss 0x2b 43
ds 0x2b 43
es 0x2b 43
fs 0x0 0
gs 0x63 99
I think the answer for 0x14(%esp,%ebx,4)
is (%ebx*4)+%esp+0x14
but what I got was 0xffffcd82
and I don't know what address that is from the registers. Can someone explain to be what value I'm supposed to put in %eax
?
Yes, you are right that 0x14(%esp,%ebx,4)
is at&t syntax for (%ebx*4)+%esp+0x14
. As such, the address is 0xffffcd88
. You can have gdb calculate that for you using p/x $ebx*4+$esp+0x14
. The add
instruction will fetch the 4 byte integer in memory at that address and add it to whatever is already in %eax
. You can check the memory contents in gdb using for example x/d 0xffffcd88
.
PS: you can switch gdb to use intel syntax which should be easier to read using set disassembly-flavor intel
.
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