How to reverse a byte

Question

I am currectly working on a project and it happens that I have to reverse the order of a byte. I am currently using AVR Studio Mega32 Microcontroller.

For example:

0000 0001 becomes 1000 0000
0001 0110 becomes 0110 1000
1101 1001 becomes 1001 1011

To start I have this:

ldi r20,0b00010110

What is the easiest way to reverse the byte so that r20 becomes 01101000?

Answer 1

Here's a snippet - it's written for the GNU toolchain (avr-gcc, binutils, avr-libc, etc) - but it should be easy to adapt:

static inline __attribute__ ((always_inline))
uint8_t avr_reverse_byte (uint8_t x)
{
    x = ((x & 0x55) << 1) | ((x & 0xaa) >> 1);
    x = ((x & 0x33) << 2) | ((x & 0xcc) >> 2);

    /* x = ((x & 0x0f) << 4) | ((x & 0xf0) >> 4); */

    __asm__ ("swap %0" : "=r" (x) : "0" (x)); /* swap nibbles. */

    return x;
}

So, not much of an improvement over the 'C' code, except for the final hi-lo nibble swap implemented with the swap instruction.

Answer 2

I can't provide AVR code just now. But the general bit reversing technique is following:

abcd efgh   p
badc fehg   p = ((p and 0AAh) shr 1) or ((p shl 1) and 0AAh)
dcba hgfe   p = ((p and 033h) shr 2) or ((p shl 2) and 033h)
hgfe dcba   p = ((p and 00Fh) shr 4) or ((p shl 4) and 0F0h)

Answer 3

Another easy way is to use the carry flag:

Repeat 8x:

lsl r20 ; shift one bit into the carry flag
ror r0  ; rotate carry flag into result 

(Input in r20, output in r0, content of r20 destroyed; registers may be changed freely.)

This uses 16 instructions @ 2 bytes, 1 cycle each = 32 bytes of program memory and 16 cycles to reverse one byte when completely 'unrolled'. Wrapped in a loop, the code size can be reduced but execution time will increase.