LSB to MSB bit reversal on ARM

Question

I need to reverse an YUV image with each byte in LSB instead of MSB. I have read Best Algorithm for Bit Reversal ( from MSB->LSB to LSB->MSB) in C but I would like to do something that is ARM-optimized.

int8 *image;
for(i = 0; i < size; i++) {
    image[i] = reversebit8(image[i]); //Use the lookup mechanism
}

As I control the image format (two-byte YUYV or any permutation), I would be OK to reverse 16 bits:

int16 *image;
for(i = 0; i < size / 2; i++) {
    image[i] = reversebit16(image[i]);
}

Image goes from YUYV LSB to UYVY MSB. Or even in 32 bits:

int32 *image;
for(i = 0; i < size / 4; i++) {
    image[i] = reversebit32(image[i]);
}

Image goes from YUYV LSB to VYUY MSB.

Question: How can I do that in an optimized way for ARM? Neon is good too.

I think that those instructions http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.kui0100a/armasm_cihjgdid.htm could be useful.

Answer 1

The ARM RBIT instruction does what you want. Write a loop that calls this on every aligned 4-byte value.

A8.6.134 RBIT
Reverse Bits reverses the bit order in a 32-bit register. Encoding T1 ARMv6T2, ARMv7

RBIT<c> <Rd>,<Rm>

if ConditionPassed() then
   EncodingSpecificOperations(); bits(32) result;
   for i = 0 to 31 do
      result<31-i> = R[m]<i>;
   R[d] = result;

Answer 2

As I mentioned it in my question, those ARM instructions (http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.kui0100a/armasm_cihjgdid.htm) can do the trick:

int32 *image;
for (i = 0; i < size / 4; h++) {
    asm("rbit %1,%0" : "=r" (image[i]) : "r" (image[i]));
    asm("rev %1,%0" : "=r" (image[i]) : "r" (image[i]));
}

rbit reverses the 32 bits, bit by bit. rev reverses the 32 bits, byte by byte. In fine, each byte is reversed independently. I'm still wondering if there is a better syntax or a better way to do this.