rookie 68000 programming session results

I left a list using a load of instructions here that was used during a rookie programming session, so, let’s see how well I did.


Results

          move.w    #$0010,d0
          mulu.w    #$0003,d0
          neg.w     d0
          ext.l     d0
          asr.l     #$01,d0
          ror.w     #$01,d0
          ext.l     d0
          divu.w    #$0002,d0

d0 starts with 00000000.

          move.w    #$0010,d0

d0 now contains 00000010.

          mulu.w    #$0003,d0

0010 x 0003 = 00000030

d0 now contains 00000030.

          neg.w     d0

0030 is changed to -0030 (FFD0).  Now d0 contains 0000FFD0.

          ext.l     d0

The word FFD0 inside d0 is extended to long-word, currently d0 contains:

0000 0000 0000 0000 1111 1111 1101 0000

The MSB of FFD0 is 1 (set), so d0 is extended to:

1111 1111 1111 1111 1111 1111 1101 0000

Now d0 contains FFFFFFD0.

          asr.l     #$01,d0

d0 contains FFFFFFD0 in binary:

1111 1111 1111 1111 1111 1111 1101 0000

Arithmetically shifting right by 1 will give us:

> 1111 1111 1111 1111 1111 1111 1110 1000 >

Now d0 contains FFFFFFE8.

          ror.w     #$01,d0

The word FFE8 of d0 is now rotated right by 01 bit:

> 0111 1111 1111 0100 v
^ < < < < < < < < < < <

Now d0 contains FFFF7FF4.

          ext.l     d0

The word 7FF4 inside d0 is extended to long-word, currently d0 contains:

1111 1111 1111 1111 0111 1111 1111 0100

The MSB of 7FF4 is 0 (clear), so d0 is extended to:

0000 0000 0000 0000 0111 1111 1111 0100

Now d0 contains 00007FF4.

          divu.w    #$0002,d0

00007FF4 ÷ 0002 = 3FFA r0000

Now d0 contains 00003FFA.

And there is my answer, d0 contains 00003FFA afterwards.  Please remember, I don’t expect myself  to work these all out in my head, if I’d prefer to use a calculator, that’s fine. -APU