为什么我在 CUDA 程序中得到 "Unspecified Launch failure"，乘以 2 个矩阵

Question

我是 CUDA 新手。当我乘以 1024x1024 矩阵并启动内核时：

multiplyKernel << <dim3(32,32, 1), dim3(32, 32, 1) >> >(dev_c, dev_a, dev_b, size);

但是当我将 2048 x 2048 矩阵与 dim3(64,64,1) 我收到此错误：

cudaDeviceSynchronize returned error code 4 after launching addKernel!
unspecified launch failure

通过修改代码，我认为错误在这个语句中

result += a[row * size + ind] * b[col + size * ind];

在

部分

b[col+size*ind]

如果我把它去掉，我不会得到内核启动错误（显然只是错误的答案）。我不知道出了什么问题。任何建议将不胜感激。我正在使用 Visual Studio 2013。我正在使用调试器，但这并不能帮助我找到错误。

这似乎是一个类似的问题： cudaDeviceSynchronize returned error code 4 after launching

非常感谢，这是代码：

cudaError_t multiplyWithCuda(int *c, const int *a, const int *b, unsigned int size); 
__global__ void multiplyKernel(int *c, const int *a, const int *b, unsigned     int size)
 {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;

if (row > size || col > size) return;

// target field in 1-D
int z = row * size + col;


int result = 0;
for (int ind = 0; ind < size  ; ++ind) {

    result += a[row * size + ind] * b[col + size * ind];

}
c[z] = result;

}

int main(){


const int sizeMatrix = 2048;
int* a = new int[sizeMatrix * sizeMatrix];
int* b = new int[sizeMatrix * sizeMatrix];
int* c = new int[sizeMatrix * sizeMatrix];



for (int i = 0; i < sizeMatrix * sizeMatrix; i++) {
    a[i] = rand() % 2;
    b[i] = rand() % 2;
}
cudaError_t cudaStatus = multiplyWithCuda(c, a, b, sizeMatrix);
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "addWithCuda failed!");
    return 1;
}


cudaStatus = cudaDeviceReset();
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaDeviceReset failed!");
    return 1;
}

return 0;
}


cudaError_t multiplyWithCuda(int *c, const int *a, const int *b, unsigned     int size)
{
int *dev_a ;
int *dev_b;
int *dev_c;
cudaError_t cudaStatus;




// Choose which GPU to run on, change this on a multi-GPU system.
cudaStatus = cudaSetDevice(0);
fprintf(stdout, "device set");
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaSetDevice failed!  Do you have a CUDA-capable GPU installed?");
    goto Error;
}

// Allocate GPU buffers for three vectors (two input, one output)    .
cudaStatus = cudaMalloc((void**)&dev_c, size * size * sizeof(int));
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMalloc failed!");
    goto Error;
}
fprintf(stdout, "buffer for c allocated \n");

cudaStatus = cudaMalloc((void**)&dev_a, size * size * sizeof(int));
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMalloc failed!");
    goto Error;
}
fprintf(stdout, "buffer for a allocated \n");

cudaStatus = cudaMalloc((void**)&dev_b, size * size * sizeof(int));
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMalloc failed!");
    goto Error;
}
fprintf(stdout, "buffer for b allocated \n");


// Copy input vectors from host memory to GPU buffers.
cudaStatus = cudaMemcpy(dev_a, a, size * size * sizeof(int), cudaMemcpyHostToDevice);
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMemcpy failed!");
    goto Error;
}
fprintf(stdout, "cudaMemcpy a done \n");


cudaStatus = cudaMemcpy(dev_b, b, size * size * sizeof(int), cudaMemcpyHostToDevice);
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMemcpy failed!");
    goto Error;
}
fprintf(stdout, "cudaMemcpy b done\n");

fprintf(stdout, "about to launch kernel \n");


// Launch a kernel on the GPU with one thread for each element.
multiplyKernel << <dim3(64,64, 1), dim3(32, 32, 1) >> >(dev_c, dev_a, dev_b, size);


fprintf(stdout, "kernel launched\n");


// Check for any errors launching the kernel
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess) {
    ; fprintf(stderr, "addKernel launch failed: %s\n", cudaGetErrorString(cudaStatus));
    goto Error;
}

// cudaDeviceSynchronize waits for the kernel to finish, and returns
// any errors encountered during the launch.
cudaStatus = cudaDeviceSynchronize();
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaDeviceSynchronize returned error code %d after launching addKernel!\n", cudaStatus);
    fprintf(stderr, " %s\n", cudaGetErrorString(cudaStatus));

    goto Error;
}

// Copy output vector from GPU buffer to host memory.
cudaStatus = cudaMemcpy(c, dev_c, size * size * sizeof(int), cudaMemcpyDeviceToHost);
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMemcpy failed!");
    goto Error;
}



Error:
  cudaFree(dev_c);
  cudaFree(dev_a);
  cudaFree(dev_b);

  return cudaStatus;
}

Answer 1

在 Windows 上，我右键单击了系统托盘中的 NSight 监视器图标。在那里我选择了选项>一般。我们看到 WDDM TDR 延迟。它是 2，我将它增加到 10。然后，我再次运行我的程序，它运行良好。这是根据罗伯特的 link（见上文） http://http.developer.nvidia.com/NsightVisualStudio/2.2/Documentation/UserGuide/HTML/Content/Timeout_Detection_Recovery.htm

为什么我在 CUDA 程序中得到 "Unspecified Launch failure"，乘以 2 个矩阵

Why do I get "Unspecified Launch failure" in CUDA program, multiplying 2 matrices

cuda

matrix

nsight