问 Cuda GPUassert:遇到非法内存访问

EN

当使用 __device__ 变量时，它们本质上处于全局范围，我们不将这些变量作为内核参数传递。您可以在内核代码中直接使用这些变量，而不必为它们设置内核参数。

如果您对代码进行了以下更改，它将无错误地运行：

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <stdlib.h>
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char* file, int line, bool abort = true)
    if (code != cudaSuccess)
        fprintf(stderr, "GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
        if (abort) exit(code);
#define M 3
#define N 3
#define K 3
using namespace std;
__device__ double A_dev[M * K];
__device__ double B_dev[K * N];
__device__ double C_dev[M * N];
__global__ void gemm(int m, int n, int k)
    int x = blockDim.x * blockIdx.x + threadIdx.x;
    int y = blockDim.y * blockIdx.y + threadIdx.y;
    int i = x * n + y;
    double sum = 0.0;
    for (int j = 0; j < k; j++)
        sum += A_dev[x * k + j] * B_dev[n * j + y];
    C_dev[i] = sum;
    printf("The value is %f", C_dev[i]);
int main(void)
    double A_h[M * K];
    double B_h[K * N];
    double C_h[M * N];
    for (int i = 0; i < M*K; i++)
        A_h[i] = (double)i;
        B_h[i] = (double)i;
        C_h[i] = 0.0;
    gpuErrchk(cudaMemcpyToSymbol(A_dev, A_h, M * K * sizeof(double), 0, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpyToSymbol(B_dev, B_h, K * N * sizeof(double), 0, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpyToSymbol(C_dev, C_h, M * N * sizeof(double), 0, cudaMemcpyHostToDevice));
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize());
    dim3 dimGrid(1, 1, 1);
    dim3 dimBlock(3, 3, 1);
    gemm <<<dimGrid, dimBlock >>> (3, 3, 3);
    gpuErrchk(cudaPeekAtLastError());