CODEWITHSUNDEEP
assembly68000easy68k
sangam.saga
sangam.saga

Reputation: 11

how does indirect addressing work in assembly language motorola 68k

I have this code:

loop2
        move.b  (a4)+,d3      * moving the morse code array input to d3
        muls    #5,d3
        add.b   $d3(a6),d3    * moving the character in morse code array to d4
        move.b  d3,d4
        cmp.b   #dot,d4

Here, I am accessing an array starting at a4. I am taking an element of the array and multiplying it with 5 to move to the memory location where I have the desired element.

a6 represents the starting point of an array which contains certain character.

the statement $d3(a6),d3 wokrs but the code gets faulty as i know the code d3(a6),d3 should be correct but it show me an error. how should i do it ?

Upvotes: 0

Views: 643

Answers (2)

lvd
lvd

Reputation: 812

In add.b (a6,d3),d3 you probably should declare size of d3 used for address generation: either (a6,d3.w) or (a6,d3.l), depending on what you need. In general case, after a multiplication you have full 32bit result and should use d3.l. Only if you are sure that the result of multiplication fits 16bit you can use d3.w.
If you write code specifically for 68020+ (020,030,040,060), also consider using scaled addressing mode like (a6,d3.w*n), where n is one of 1, 2, 4, 8.

Upvotes: 0

Drew McGowen
Drew McGowen

Reputation: 11706

What's happening is since $ is for specifying hex constants, and d3 is technically a valid hex constant, it's using a6 + 0xd3 as the memory address. What you're trying to accomplish uses different syntax, namely both registers are in the parentheses: (a6,d3)

So the correct statement is:

add.b   (a6,d3),d3

Upvotes: 0

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