26864

Swapping positions (HC11)

<h3>Question</h3>

I'm just playing with my MC 68HC11; in C i can perform a simple byte swap by doing something like this:

swapped = ((num>>24)&0xff) | // move byte 3 to byte 0 ((num<<8)&0xff0000) | // move byte 1 to byte 2 ((num>>8)&0xff00) | // move byte 2 to byte 1 ((num<<24)&0xff000000); // byte 0 to byte 3

But now i want to achieve something a little harder using assembly code:

I created an ARRAY and added some values (using little endian logic). I want to read that ARRAY and swap all the values into big endian logic and store them inside "BIGENDIAN". I was thinking something like this:

RWM EQU $0 ROM EQU $C000 RESET EQU $FFFE ORG RWM BIGENDIAN RMB 16 ORG ROM Main: END BRA END ARRAY DW $0124,$FEEB,$0011,$0070,$ABEF,$074B,$8004,$8080 ORG RESET DW Main

I tried but it did not work properly.


<h3>Answer1:</h3>

DW creates 16-bit words. (Your C example is for 32-bit words.)

For 16-bit one possibility is this:

RAM equ $0 ROM equ $C000 Vreset equ $FFFE org RAM BIGENDIAN rmb 16 org ROM ARRAY dw $0124,$FEEB,$0011,$0070,$ABEF,$074B,$8004,$8080 Start ldx #ARRAY ;X -> source ldy #BIGENDIAN ;Y -> destination Loop ldd ,x ;A = MSB, B = LSB (big endian view) staa 1,y stab ,y ; one alternative for above two instructions ; pshb ; tab ; pula ; std ,y ldab #2 ;B = word size abx ;X -> next word aby ;Y -> next word cpx #ARRAY+::ARRAY blo Loop Done bra * org Vreset dw Start

If you meant 32-bit words, so two 16-bit should be considered one word, I'll revise.

来源:https://stackoverflow.com/questions/62296789/swapping-positions-hc11

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