CUDA编程未处理的异常和堆栈溢出[重复]

Question

我正在尝试比较并行添加向量比非并行添加更快。为了使并行添加更快，我需要大数字。当我的 N 为 10000 时，我仍然可以运行它。但是，当我的 N 为 100000 时，我得到

basicCuda.exe 中 0x00D25B89 处未处理的异常：0xC00000FD：堆栈 溢出（参数：0x00000000、0x002F2000）

我该如何解决这个问题？

#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <iostream>
#include <stdio.h>
#include <time.h>
#include <chrono>
cudaError_t addWithCuda(int *c, const int *a, const int *b, uint64_t N);

__global__ void addKernel(int *c, const int *a, const int *b, uint64_t N)
{
    int i = threadIdx.x + blockIdx.x * blockDim.x;
    if (i < N) {
        c[i] = a[i] * b[i];
    }
}


void randomizeArr(int arr[],  int size) {

    unsigned int randNum;
    srand(time(NULL));
    for (int i = 0; i < size; i++) {
        randNum = rand() % 100 + 1;
        arr[i] = randNum;
    }
}

void addWithCPU(int c[], int a[], int b[], int size) {
    for (int i = 0; i < size; i++) {
        c[i] = a[i] + b[i];
    }
}

#define N (10000) // Number of elements each array has
#define M 1024 // 512 Threads Per Block
int main()
{
    const uint64_t arSize = N;
    int a[arSize] = { 0 };
    int b[arSize] = { 0 };
    randomizeArr(a, arSize);
    randomizeArr(b, arSize);
    int c[arSize] = { 0 };
    int d[arSize] = { 0 };

    // Add vectors in parallel.
    int iterations = 100;
    cudaError cudaStatus;
    auto begin = std::chrono::high_resolution_clock::now();
    for (uint32_t i = 0; i < iterations; ++i)
    {
        cudaStatus = addWithCuda(c, a, b, arSize);
    }
    auto end = std::chrono::high_resolution_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin).count();
    std::cout << "Parallel : " << duration / iterations << "ns." << std::endl;

    // Add vectors NOT in parallel.
    auto begin2 = std::chrono::high_resolution_clock::now();
    for (uint32_t i = 0; i < iterations; ++i)
    {
        addWithCPU(d, a, b, arSize);
    }
    auto end2 = std::chrono::high_resolution_clock::now();
    auto duration2 = std::chrono::duration_cast<std::chrono::nanoseconds>(end2 - begin2).count();
    std::cout << "Not Parallel : " << duration2 / iterations << "ns." << std::endl;


    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "addWithCuda failed!");
        std::getchar();
        return 1;
    }


    // 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;
    }
    std::getchar();

    return 0;
}

// Helper function for using CUDA to add vectors in parallel.
cudaError_t addWithCuda(int *c, const int *a, const int *b, uint64_t size)
{
    int *dev_a = 0;
    int *dev_b = 0;
    int *dev_c = 0;
    cudaError_t cudaStatus;

    // Choose which GPU to run on, change this on a multi-GPU system.
    cudaStatus = cudaSetDevice(0);
    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 * sizeof(int));
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaMalloc failed!");
        goto Error;
    }

    cudaStatus = cudaMalloc((void**)&dev_a, size * sizeof(int));
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaMalloc failed!");
        goto Error;
    }

    cudaStatus = cudaMalloc((void**)&dev_b, size * 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, size * sizeof(int), cudaMemcpyHostToDevice);
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaMemcpy failed!");
        goto Error;
    }

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

    // Launch a kernel on the GPU with one thread for each element.
    addKernel<<<(N + M - 1)/ M, M>>>(dev_c, dev_a, dev_b, 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);
        goto Error;
    }

    // Copy output vector from GPU buffer to host memory.
    cudaStatus = cudaMemcpy(c, dev_c, 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

所有数组：

int a[arSize]
int b[arSize]
int c[arSize]
int d[arSize]

在函数“main”中的堆栈上创建。给定 arSize = 100000 和 sizeof(int) = 4，您要求分配 1600000 字节 (1.5 MB)，这可能需要使用编译器或操作系统参数进行调整以允许这么大的堆栈。

相反，您可以使用new 分配内存：

int* a = new int[arSize]();

（注意所有数组值都会被初始化为0，另见c++ initial value of dynamic array）