如何在CUDA中安全地将全局内存中的数据加载到共享内存中？

Question

My kernel:

__global__ void myKernel(float * devData, float * devVec, float * devStrFac,
int Natom, int vecNo) {

extern __shared__ float sdata[];
int idx = blockIdx.x * blockDim.x + threadIdx.x;

float qx=devVec[3*idx];
float qy=devVec[3*idx+1];
float qz=devVec[3*idx+2];
__syncthreads();//sync_1

float c=0.0,s=0.0;
for (int iatom=0; iatom<Natom; iatom += blockDim.x) {
    float rtx = devData[3*(iatom + threadIdx.x)];//tag_0
    float rty = devData[3*(iatom + threadIdx.x)+1];
    float rtz = devData[3*(iatom + threadIdx.x)+2];
    __syncthreads();//sync_2
    sdata[3*threadIdx.x] = rtx;//tag_1
    sdata[3*threadIdx.x + 1] = rty;
    sdata[3*threadIdx.x + 2] = rtz;
    __syncthreads();//sync_3

    int end_offset=  min(blockDim.x, Natom - iatom);

    for (int cur_offset=0; cur_offset<end_offset; cur_offset++) {
        float rx = sdata[3*cur_offset];
        float ry = sdata[3*cur_offset + 1];
        float rz = sdata[3*cur_offset + 2];
        //sync_4  
        float theta = rx*qx + ry*qy + rz*qz;

        theta = theta - lrint  (theta);
        theta = theta * 2 * 3.1415926;//reduce theta to [-pi,pi]

        float ct,st;
        sincosf(theta,&st,&ct);

        c += ct;
        s += st;
    }

}

devStrFac[idx] += c*c + s*s;
}

why "__syncthreads()" labeled sync_2 is needed? Without sync_2, sdata[] get wrong numbers and I get wrong results. Line "tag_1" use the results of line "tag_0", so in my mind sync_2 is no need. Where do I wrong? If due to disorderd instruction executing, I should put a __syncthreads() in line "sync_4"?

Answer 1

Consider one warp of the thread block finishing the first iteration and starting the next one, while other warps are still working on the first iteration. If you don't have __syncthreads at label sync2 , you will end up with this warp writing to shared memory while others are reading from that shared memory, which is race condition.

You might move this __syncthreads() at label sync2 to the end of the outer loop for the sake of clarity.

"cuda-memcheck --tool racecheck" should tell you where the problem is.