How can I simulate the LEA instruction with MOVE?

Question

How can I simulate the LEA instruction with MOVE? is this possible? I've used x86 assembly and the MOV instruction just copy adddress of if it's a symbol (not sure if it's assembly-dependent). But in 6800 MOVE looks like behave different, e.g., in the following instructions D1 and A1 has different values. How does LEA instruction work exactly? and how can I simulate? i.e., do what the LEA do without use it. I think that with MOVE and some others. Also, what does, de facto, MOVE is putting in D1 it doesn't seem the hex values of label.

  MOVE T(PC),D1
  LEA T,A1

Nothing changed. Same behavior. Same values in registers: one different from other. — The Mask, Nov 28 '13 at 20:09
Oops. That should be `MOVEA.L #T,A1`. Or to put the address in a data register: `MOVE.L #T,D1` (note that `MOVE` affects some of the condition flags, while `MOVEA` and `LEA` don't). — Michael, Nov 28 '13 at 20:16
Also, `MOVE` itself default to what? `MOVE.L` or `MOVE.W` or `MOVE.B`? — The Mask, Nov 28 '13 at 20:20

score 3 · Accepted Answer · answered Nov 28 '13 at 20:31

3

For an address register:

MOVEA.L #T,A<n>

For a data register:

MOVE.L #T,D<n>

Note that MOVE will update some of the condition flags, which neither LEA nor MOVEA does.

answered Nov 28 '13 at 20:31

Michael

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I will use `MOVEA` instead of. Thanks very much! – The Mask Nov 28 '13 at 20:35

How can I simulate the LEA instruction with MOVE?

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