x86 instruction pointer help please!

sakatari's Avatar, Join Date: Oct 2010
Newbie Member
Hi, I'm currently working on understanding the x86 assembly language. I am currently stuck on understanding this instruction:

mov 0x28e89c (%rip), %eax

I know that rip is the instruction pointer that holds the address of the next instruction and that eax is the destination for whatever 0x28e89c (%rip) is. I assume that the 0x28e89c is an offset of some sort but I don't know how to interpret it. If anyone could help it would be much appreciated, thanks!
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lionaneesh's Avatar, Join Date: Mar 2010
Invasive contributor
Quote:
Originally Posted by sakatari View Post
Hi, I'm currently working on understanding the x86 assembly language. I am currently stuck on understanding this instruction:

mov 0x28e89c (%rip), %eax

I know that rip is the instruction pointer that holds the address of the next instruction and that eax is the destination for whatever 0x28e89c (%rip) is. I assume that the 0x28e89c is an offset of some sort but I don't know how to interpret it. If anyone could help it would be much appreciated, thanks!
EIP is the instruction pointer not RIP...
Maybe this is a typo...

0x28e89c is the offset to the address pointed by eip

Eg :-

if EIP points to 0xDEADBEEF

The the instruction will point to :-

Code:
0xDEADBEEF + 0x28e89c = 0xDED6A78B
I hope it helps..