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Hi all,
I've been trying to understand how an assembler assembles the JMP command. Consider the following code:
The assembler turns this into the following hex:
The FC at the end corresponds to a displacement of -4. I would have thought that the displacement would only need to be -3 (
Thanks
--Amr
I've been trying to understand how an assembler assembles the JMP command. Consider the following code:
AGAIN:
OUT 0,AL
JMP AGAINThe assembler turns this into the following hex:
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
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Maybe because PC is already pointing at the first byte of the next instruction, rather than the last data byte of the JMP instruction.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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Originally Posted by amrbekhit 
Hi all,
I've been trying to understand how an assembler assembles the JMP command. Consider the following code:
AGAIN: OUT 0,AL JMP AGAIN
The assembler turns this into the following hex:
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
Yeah, the program counter automatically points to the next instruction before the actual execution, so the jump back to FC is counted as well.
Last edited by Colin Mac : May 4th, 2007 at 12:26 pm.
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