How to load register with predefined bytes in Assembly

Question

Using ARM assembly, I want to load R0 with predefined bytes A, B, C, D. In the .data section I have them defined as:

A:    .byte    0xFF
B:    .byte    0xAA
C:    .byte    0x88
D:    .byte    0x77

I want R0 to be FFAA8877 when all is said and done. Not sure I should be using .byte or .word or even something else for A, B, C, D.

EDIT: Here is what I'm trying to do with R0:

@on entry: R0 holds the word to be swapped
@on exit: R0 holds the swapped word, R1 is destroyed
@R0 = A, B, C, D
byteswap:
    EOR R1, R0, R0, ROR #16
    BIC R1, R1, #0xFF0000
    MOV R0, R0, ROR #8
    EOR R0, R0, R1, LSR #8

If you really need to have them in the datasection for some reason, just define it as MyValue: .word 0xFFAA8877 and later in the code use ldr r0, =MyValue and ldr r0, [r0] The first will load the address the second will load from that address. — Nico Erfurth
– Nico Erfurth, Commented Oct 6, 2013 at 21:58

old_timer · Accepted Answer · 2013-10-07 12:48:34Z

1

Here is a shortcut.

ldr r0,=0xFFAA8877

Or you can do the same thing manually...

ldr r0,my_number
...
my_number: .word 0xFFAA8877

edited Oct 7, 2013 at 12:48

answered Oct 7, 2013 at 3:17

old_timer

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Comments

Jitesh · Accepted Answer · 2013-10-06 22:45:04Z

0

Assuming you are running little-endian ARM, here is how you can do it:

ldr r0, =A
ldr r1, [r0]
rev r0, r1

Note that the rev instruction does the conversion from little-endian to big-endian. Note that the rev instruction is available only armv6+.

If you are running big-endan ARM, just skip the rev instruction and you'll be fine :)

Edit: rev instruction was incorrect.

edited Oct 6, 2013 at 22:45

answered Oct 6, 2013 at 22:39

Jitesh

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1 Comment

Alex Mohr Over a year ago

The purpose of this is to swap the bytes of R0 between little-endian and big-endian. I'll add the instructions to the OP

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