x86 instruction pointer help please!
 

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!

Re: x86 instruction pointer help please!
 

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 :-

I hope it helps..