How is this 68K instruction addressed?

[pasky13]

027218: move.l ($22,PC,D0.w), D0

Code: Select all

A0=0003B854, A1=0090C280, A2=00085410, A3=00179042, A4=FFFF8800, A5=00FF8000, A6=00FF8400, A7=00FF06D2, D0=00000058, D1=0000004C, D2=0009EE00, D3=FF00F002, D4=F000F000, D5=00000007, D6=0000FFFF 027218: move.l  ($22,PC,D0.w), D0
A0=0003B854, A1=0090C280, A2=00085410, A3=00179042, A4=FFFF8800, A5=00FF8000, A6=00FF8400, A7=00FF06D2, D0=00013333, D1=0000004C, D2=0009EE00, D3=FF00F002, D4=F000F000, D5=00000007, D6=0000FFFF 02721C: move.b  ($a,A6), D1

I thought it would be the pointer at 0x22 + 0x27218 + 0x58 = 0x27292 but the address that gets written to D0 does not match up with the pointer in that address. How is this calculated? Thanks.

[pasky13]

Surely someone has run into this type of addressing before? =/

[Pugsy]

I can't remember, the more obscure instructions like this which you often see more of in compiled code can be worked out by looking at the memory at the time or by using a large read watchpoint, without the name of the game or the bytes needed to implement the instruction I'm not going to look.

[pasky13]

It's Vsav, moving anakaris forward or backward.

[Pugsy]

A quick solution....you are getting confused because it uses encrypted memory so the memory viewer window is showing the encrypted data.

In the example above:-

0x22 + 0x27218 + 0x58 = 0x27292

this is correct...BUT you should look at what's in 27292 in the disassembly window rather than the memory viewer window to see the unencrypted values rather than the encrypted values.

[pasky13]

I see, thanks.