How to mask bytes in ARM assembly?

Question

i have got an 32bit (hexadecimal)word 0xaabbccdd and have to swap the 2. and the 3. byte. in the end it should look like 0xaaccbbdd

how can i "mask" the 2nd and the 3rd byte to first load them up to register r1 and r2 and the swap them.. i also know that i have to work with lsl and lsr commands but dont know how to start.

sorry for my bad english.hope anyone could help me out!

regards, sebastian

Answer 1

Back in the day we used to rely heavily on EOR for this kind of trickery.

You can do it in 4 cycles.

First off, we need the fact that: A ^ (A^B) = B

We start with 0xAABBCCDD, and we want 0xAACCBBDD. To get there, we need 0x00EEEE00^0xAABBCCDD, where EE = BB^CC.

Now, we need a few cycles to build 00EEEE00:

eor     r1,r0,r0,lsr #8
and     r1,r1,#0xFF00
orr     r1,r1,r1,lsl #8
eor     r0,r0,r1

In c:

t=x^(x>>8);
t=t&0xFF00;
t=t|(t<<8);
x^=t;

After each line, the result calculated is: starting with: AABBCCDD

eor  XXXXEEXX
and  0000EE00
orr  00EEEE00
eor  AACCBBDD

This will work on any 32bit ARM core.