Pointers and Indexes in Intel 8086 Assembly

Question

I have a pointer to an array, DI.

Is it possible to go to the value pointed to by both DI and another pointer?

eg:

mov bl,1           
mov bh,10
inc [di+bl]
inc [di+bh]

And, on a related note, is there a single line opcode to swap the values of two registers? (In my case, BX and BP?)

Answer 1

For 16-bit programs, the only supported addressing forms are:

[BX+SI]
[BX+DI]
[BP+SI]
[BP+DI]
[SI]
[DI]
[BP]
[BX]

Each of these may include either an 8- or 16-bit constant displacement.

(Source: Intel Developer's Manual volume 2A , page 38)

The problem with the example provided is that bl and bh are eight-bit registers and cannot be used as the base pointer. However, if you set bx to the desired value then inc [di+bx] (with a suitable size specifier for the pointer) is valid.

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As for swapping "the high and low bits of a register," J-16 SDiZ's suggestion of ror bx, 8 is fine for exchanging bl and bh (and IIRC, it is the optimal way to do so). However, if you want to exchange bit 0 of (say) bl with bit 7 of bl , you'll need more logic than that.

Answer 2

DI is not a pointer, it is an index.

You can you ROR BX, 8 to rotate a lower/higher byte of a register.