PyCUDA 和固定内存的点积

Question

我目前正在使用 PyCUDA 开发带有固定内存的点积。我对大数组有疑问。

我正在与：

英伟达 GTX 1060
CUDA 9.1
PyCUDA 2017.1.1

密码是：

#!/usr/bin/env python

import numpy as np
import argparse
import math
import pycuda.autoinit
import pycuda.driver as drv
from pycuda.compiler import SourceModule

from time import time

dot_mod = SourceModule("""
__global__ void full_dot( double* v1, double* v2, double* out, int N ) {
    __shared__ double cache[ 1024 ];
    int i = blockIdx.x * blockDim.x + threadIdx.x;
    cache[ threadIdx.x ] = 0.f;
    while( i < N ) {
        cache[ threadIdx.x ] += v1[ i ] * v2[ i ];
        i += gridDim.x * blockDim.x;
    }
    __syncthreads(); // required because later on the current thread is accessing
                     // data written by another thread    
    i = 1024 / 2;
    while( i > 0 ) {
        if( threadIdx.x < i ) cache[ threadIdx.x ] += cache[ threadIdx.x + i ];
        __syncthreads();
        i /= 2; //not sure bitwise operations are actually faster
    }

#ifndef NO_SYNC // serialized access to shared data; 
    if( threadIdx.x == 0 ) atomicAdd( out, cache[ 0 ] );
#else // no sync, what most likely happens is:
      // 1) all threads read 0
      // 2) all threads write concurrently 16 (local block dot product)
    if( threadIdx.x == 0 ) *out += cache[ 0 ];
#endif                

}
""")


def main(args):
    dot = dot_mod.get_function("full_dot")
    N = args.number
    BLOCK_SIZE = 1024
    BLOCKS = int(math.ceil(N/BLOCK_SIZE))
    THREADS_PER_BLOCK = BLOCK_SIZE

    # Time use of pinned host memory:
    x = drv.aligned_empty((N), dtype=np.float64, order='C')
    x = drv.register_host_memory(x, flags=drv.mem_host_register_flags.DEVICEMAP)
    x_gpu_ptr = np.intp(x.base.get_device_pointer())

    # Time use of pinned host memory:
    y = drv.aligned_empty((N), dtype=np.float64, order='C')
    y = drv.register_host_memory(y, flags=drv.mem_host_register_flags.DEVICEMAP)
    y_gpu_ptr = np.intp(y.base.get_device_pointer())

    # Time use of pinned host memory:
    z = drv.aligned_empty((1), dtype=np.float64, order='C')
    z = drv.register_host_memory(z, flags=drv.mem_host_register_flags.DEVICEMAP)
    z_gpu_ptr = np.intp(z.base.get_device_pointer())

    z[:] = np.zeros(1)
    x[:] = np.zeros(N)
    y[:] = np.zeros(N)

    x[:] = np.random.rand(N)
    y[:] = x[:] 
    x_orig = x.copy()
    y_orig = y.copy()

    start = time()
    dot(x_gpu_ptr, y_gpu_ptr, z_gpu_ptr, np.uint32(N), block=(THREADS_PER_BLOCK, 1, 1), grid=(BLOCKS,1))
    times = time()-start
    print "Average kernel GPU dot product execution time with pinned memory:   %3.7f" % np.mean(times)

    start = time()
    ydot=np.dot(x_orig,y_orig)
    times = time()-start
    print "Average numpy dot product execution time:   %3.7f" % np.mean(times)

    print N,ydot,z[0]

if __name__ == "__main__":

    parser = argparse.ArgumentParser(description=' ')
    parser.add_argument('-n', dest='number', type=long, help="Number of samples ", required=True)

    args = parser.parse_args()

    main(args)

我编写的这段代码在样本数组的大小大约小于 1024*12 时运行良好，但对于像 1024*1024 这样的大数组，值会给出错误的结果。-

➜  ./test_dot_pinned.py -n 16384
Average kernel GPU dot product execution time with pinned memory:   0.0001669
Average numpy dot product execution time:   0.0000119
16384 5468.09590706 5468.09590706
 SIZE   np.dot()    GPU-dot-pinned

➜  ./test_dot_pinned.py -n 1048576
Average kernel GPU dot product execution time with pinned memory:   0.0002351
Average numpy dot product execution time:   0.0010922
1048576 349324.532564 258321.148593
 SIZE   np.dot()    GPU-dot-pinned

谢谢大家，希望有人能帮帮我。

Answer 1

pycuda 在内核启动后不强制执行任何同步。通常，如果您在内核启动后执行设备-> 主机数据副本，该操作将强制同步，即它会强制内核完成。

但是你的代码中没有这样的同步。由于您使用的是固定内存，因此随着内核执行时间的增长（由于工作量更大），最终当您打印出 z[0] 时，您只会得到部分结果，因为此时内核尚未完成。

这样做的一个副作用是您的内核时间测量不准确。

您可以通过在完成时间测量之前强制内核完成来解决这两个问题：

dot(x_gpu_ptr, y_gpu_ptr, z_gpu_ptr, np.uint32(N), block=(THREADS_PER_BLOCK, 1, 1), grid=(BLOCKS,1))
#add the next line of code:
drv.Context.synchronize()
times = time()-start

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