How to pack bits (efficiently) in CUDA?

2
votes

I have an array of bytes where each byte is either 0 or 1. Now I want to pack these values into bits, so that 8 original bytes occupy 1 target byte, with original byte 0 going into bit 0, byte 1 into bit 1, etc. So far I have the following in the kernel:

const uint16_t tid = threadIdx.x;
__shared__ uint8_t packing[cBlockSize];

// ... Computation of the original bytes in packing[tid]
__syncthreads();

if ((tid & 4) == 0)
{
    packing[tid] |= packing[tid | 4] << 4;
}
if ((tid & 6) == 0)
{
    packing[tid] |= packing[tid | 2] << 2;
}
if ((tid & 7) == 0)
{
    pOutput[(tid + blockDim.x*blockIdx.x)>>3] = packing[tid] | (packing[tid | 1] << 1);
}

Is this correct and efficient?

2
c++parallel-processingcudabit-packing
This can't ever work. That is a memory race. There is not such thing as parallel bit sized transactions in CUDA - talonmies
@talonmies, I thought there is no race because threads handling the same byte belong to the same warp. - Serge Rogatch
Being in the same warp is no guarantee of safety, No two threads can modify the same byte simultaneously without causing a race - talonmies
@talonmies, Where are threads modifying the same byte? - Serge Rogatch
I meant that they handle the same byte sequentially between ifs, but they don't modify the same byte in parallel. - Serge Rogatch

2 Answers

8
votes

The __ballot() warp-voting function comes quite handy for this. Assuming that you can redefine pOutput to be of uint32_t type, and that your block size is a multiple of the warp size (32):

unsigned int target = __ballot(packing[tid]);
if (tid % warpSize == 0) {
    pOutput[(tid + blockDim.x*blockIdx.x) / warpSize] = target;
}

Strictly speaking, the if conditional isn't even necessary, as all threads of the warp will write the same data to the same address. So a highly optimized version would just be

pOutput[(tid + blockDim.x*blockIdx.x) / warpSize] = __ballot(packing[tid]);
1
votes

For two bits per thread, using uint2 *pOutput

int lane = tid % warpSize;
uint2 target;
target.x = __ballot(__shfl(packing[tid], lane / 2)                & (lane & 1) + 1));
target.y = __ballot(__shfl(packing[tid], lane / 2 + warpSize / 2) & (lane & 1) + 1));
pOutput[(tid + blockDim.x*blockIdx.x) / warpSize] = target;

You'll have to benchmark whether this is still faster than your conventional solution.