Understanding the 8088/86 JMP command

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

[Salem]

Maybe because PC is already pointing at the first byte of the next instruction, rather than the last data byte of the JMP instruction.

[Colin Mac]

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.