在本例中，为什么PyCUDA比CUDA快

import pycuda.driver as cuda
import pycuda.autoinit
from pycuda.compiler import SourceModule
import numpy as np

# 定义CUDA内核函数
mod = SourceModule("""
__global__ void matrix_mul(float *A, float *B, float *C, int N) {
    int row = blockIdx.y * blockDim.y + threadIdx.y;
    int col = blockIdx.x * blockDim.x + threadIdx.x;
    float sum = 0.0;
    if (row < N && col < N) {
        for (int k = 0; k < N; k++) {
            sum += A[row * N + k] * B[k * N + col];
        }
        C[row * N + col] = sum;
    }
}
""")

matrix_mul = mod.get_function("matrix_mul")

# 初始化输入矩阵
N = 1024
A = np.random.rand(N, N).astype(np.float32)
B = np.random.rand(N, N).astype(np.float32)
C = np.zeros((N, N), dtype=np.float32)

# 分配GPU内存
d_A = cuda.mem_alloc(A.nbytes)
d_B = cuda.mem_alloc(B.nbytes)
d_C = cuda.mem_alloc(C.nbytes)

# 将数据传输到GPU
cuda.memcpy_htod(d_A, A)
cuda.memcpy_htod(d_B, B)

# 设置网格和块大小
block_size = (16, 16)
grid_size = ((N + block_size[0] - 1) // block_size[0], (N + block_size[1] - 1) // block_size[1])

# 调用CUDA内核函数
matrix_mul(d_A, d_B, d_C, np.int32(N), block=block_size, grid=grid_size)

# 将结果传输回CPU
cuda.memcpy_dtoh(C, d_C)

# 验证结果
assert np.allclose(np.dot(A, B), C, atol=1e-6)