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x86 instruction pointer help please!

Discussion in 'Assembly Language Programming (ALP) Forum' started by sakatari, Oct 17, 2010.

  1. sakatari

    sakatari New Member

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    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!
     
  2. lionaneesh

    lionaneesh Active Member

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    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..
     

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