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!
Maybe this is a typo...
0x28e89c is the offset to the address pointed by eip
if EIP points to 0xDEADBEEF
The the instruction will point to :-
|All times are GMT +5.5. The time now is 02:30.|