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Strange Cudamemcpy execution time（1个答案）
9天前关闭。
我有一个在cuBLAS中执行BLAS的gemm（）子程序的程序，如下所示：

size_t m = 10000;
size_t k = 4000;
size_t n = 6000;

// h_A is an m-by-k matrix, h_B is a k-by-n matrix and h_C is a m-by-n matrix.

double *h_C;
cudaHostAlloc((void **)(&h_C), m * n * sizeof(double), cudaHostAllocDefault);

uint64_t t0 = get_timestamp_in_microsec();

CUBLAS_CHECK(cublasCreate(&cublasH));

uint64_t t1 = get_timestamp_in_microsec();

CUDA_CHECK(cudaMalloc((void **)(&d_A), sizeof(double) * m * k));
CUDA_CHECK(cudaMalloc((void **)(&d_B), sizeof(double) * k * n));
CUDA_CHECK(cudaMalloc((void **)(&d_C), sizeof(double) * m * n));

uint64_t t2 = get_timestamp_in_microsec();

CUDA_CHECK(cudaMemcpy(d_A, h_A.data(), sizeof(double) * m * k, cudaMemcpyHostToDevice));
CUDA_CHECK(cudaMemcpy(d_B, h_B.data(), sizeof(double) * k * n, cudaMemcpyHostToDevice));

uint64_t t3 = get_timestamp_in_microsec();

CUBLAS_CHECK(cublasDgemm(cublasH, CUBLAS_OP_N, CUBLAS_OP_N, m, n, k, &alpha, d_A, lda, d_B, ldb, &beta, d_C, ldc));

uint64_t t4 = get_timestamp_in_microsec();

CUDA_CHECK(cudaMemcpy(h_C, d_C, sizeof(double) * m * n, cudaMemcpyDeviceToHost));

uint64_t t5 = get_timestamp_in_microsec();

字符串
性能还可以，除了最后的cudaMemcpy()将d_C移回h_C需要大约3秒：

t1-t0: 12.929ms
t2-t1: 0.949ms
t3-t2: 53.256ms
t4-t3: 0.315ms
t5-t4: 2653.57ms

型
在搜索了SO和其他网站之后，我从malloc()迁移到cudaHostAlloc()，但上面的结果似乎没有任何变化。
我的代码中有任何问题吗？或者这已经是最佳性能了吗？