cuda错误4未指明的启动失败

Question

下面是GPU内核的代码片段：

__global_ void POCKernel(int *a) 
{
int i = threadIdx.x;
a[i] = a[i] + 1;

if (i < 1024 * 1024)
{
    double dblNewMemoryVarA[15];
    double dblNewMemoryVarB[15];
    double dblNewMemoryVarC[15];

    //double* dblNewMemoryVarA = (double*)malloc(15 * sizeof(double));
    ////memset(dblNewMemoryVarA, 0, 15 * sizeof(double));
    //double* dblNewMemoryVarB = (double*)malloc(15 * sizeof(double));
    ////memset(dblNewMemoryVarB, 0, 15 * sizeof(double));
    //double* dblNewMemoryVarC = (double*)malloc(15 * sizeof(double));
    ////memset(dblNewMemoryVarC, 0, 15 * sizeof(double));
    for (int j = 0; j < 15; j++)
    {
        dblNewMemoryVarA[j] = 0;
        dblNewMemoryVarB[j] = 0;
        dblNewMemoryVarC[j] = 0;
    }
    dblNewMemoryVarC[i] = dblNewMemoryVarA[i] + dblNewMemoryVarB[i];
    dblNewMemoryVarC[i] = dblNewMemoryVarA[i] * dblNewMemoryVarB[i];
    dblNewMemoryVarC[i] = dblNewMemoryVarA[i] - dblNewMemoryVarB[i];
    /*free(dblNewMemoryVarA);
    free(dblNewMemoryVarB);
    free(dblNewMemoryVarC);*/


}

}

对此内核的调用函数为：

int main()
{



const int arraySize = 1024 * 1024;
int* a = new int[arraySize];
int *dev_a = 0;

for (int i = 0; i < arraySize; i++)
{
    a[i] = 5;
}


cudaError_t cudaStatus;

// Choose which GPU to run on, change this on a multi-GPU system.

cudaStatus = cudaSetDevice(0);
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "CUDA failed!");
    return 1;
}

// Allocate GPU buffers for three vectors (two input, one output)    .
cudaStatus = cudaMalloc((void**)&dev_a, arraySize * sizeof(int));
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMalloc failed!");
    goto Error;
}

// Copy input vectors from host memory to GPU buffers.
cudaStatus = cudaMemcpy(dev_a, a, arraySize * sizeof(int), cudaMemcpyHostToDevice);
if (cudaStatus != cudaSuccess) {
    fprintf(stderr, "cudaMemcpy failed!");
    goto Error;
}
// Launch a kernel on the GPU with one thread for each element.
POCKernel << <4096, 256 >> >(dev_a);

// 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);
    goto Error;


    // cudaDeviceReset must be called before exiting in order for profiling and
    // tracing tools such as Nsight and Visual Profiler to show complete traces.
    cudaStatus = cudaDeviceReset();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaDeviceReset failed!");
        return 1;
    }

Error:
    cudaFree(dev_a);



    return 0;
    }
}

在cudaDeviceSynchronize上，错误代码为4-未指明的启动失败。有人能告诉我为什么我要面对这个问题吗？

Answer 1

这段代码在很多方面都很奇怪，但让我们直奔主题。下面这几行内核代码中有一个明确的问题：

int i = threadIdx.x;
...

if (i < 1024 * 1024)
{
    double dblNewMemoryVarA[15];
    double dblNewMemoryVarB[15];
    double dblNewMemoryVarC[15];
    ...
    dblNewMemoryVarC[i] = dblNewMemoryVarA[i] + dblNewMemoryVarB[i];

您正在启动每个线程包含256个线程的线程块：

POCKernel << <4096, 256 >> >(dev_a);
                    ^^^

这意味着在一个块中的所有线程中，threadIdx.x变量的范围从0到255：

int i = threadIdx.x;

在您的局部变量中，您为15个量分配了空间：

    double dblNewMemoryVarA[15];

但是您随后尝试使用i对这些数组进行索引，如前所述，它的范围最大为255：

    dblNewMemoryVarC[i] = dblNewMemoryVarA[i] + dblNewMemoryVarB[i];

因此，这将产生越界索引，这很可能导致内核启动失败。

这是不可能肯定的，因为您没有提供完整的代码，也没有说明您是如何编译的，或者您在什么环境中运行。但从代码正确性的角度来看，上述做法肯定是非法的。

我猜你是在调试模式(-G)下编译的。如果不是，我希望编译器在If测试结束后优化所有内容，因为这些代码都不会影响任何全局状态。

正如评论中所指出的，如果你正在运行这个windows，你可能会遇到一个windows WDDM超时。