Originally Posted by amrbekhit

Hi all,

I've been trying to understand how an assembler assembles the JMP command. Consider the following code:

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AGAIN:
    OUT 0,AL
    JMP AGAIN

AGAIN: OUT 0,AL JMP AGAIN

The assembler turns this into the following hex:

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E6 00 EB FC

E6 00 EB FC

The FC at the end corresponds to a displacement of -4. I would have thought that the displacement would only need to be -3 (FC->EB, EB->00, 00->E6). What is the reason for this extra move? I'm guessing this is related to the 8088 parallel fetch and execution.

Thanks

--Amr