CUDA内核函数的连续执行

以下是两个连续的CUDA核函数衔接的一种思路：

要完成的功能：1. 向量的计算computer（暂时以两向量求和为例）；2. 对结果向量求和SUM。

思路1：写1个计算的内核函数，中间结果保留，求和函数调用cublas。

#include "cuda_runtime.h"#include "device_launch_parameters.h"#include "cublas_v2.h"#include <stdio.h>#include <stdlib.h>#pragma comment(lib, "cublas.lib")cudaError_t addWithCuda(const int *a, const int *b, unsigned int size, float& sum);__global__ void addKernel(const int *a, const int *b, float *c){    int i = threadIdx.x;    c[i] = a[i] + b[i];}int main(){    const int arraySize = 5;    const int a[arraySize] = { 1, 2, 3, 4, 5 };    const int b[arraySize] = { 10, 20, 30, 40, 50 };    // Add vectors in parallel.float sum = 0.0;    cudaError_t cudaStatus = addWithCuda(a, b, arraySize, sum);    if (cudaStatus != cudaSuccess) {        fprintf(stderr, "addWithCuda failed!");        return 1;    }    printf("sum: %f\n", sum);    // 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;    }    return 0;}// Helper function for using CUDA to add vectors in parallel.cudaError_t addWithCuda(const int *a, const int *b, unsigned int size, float& sum){    int *dev_a = 0;    int *dev_b = 0;    float *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(float));    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 << <1, size >> >(dev_a, dev_b, dev_c);    // 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;    }cublasHandle_t handle;cublasStatus_t stat;stat = cublasCreate(&handle);if (stat != CUBLAS_STATUS_SUCCESS) {printf("CUBLAS initialization failed\n");exit(-1);}cublasSasum(handle, size, dev_c, 1, &sum);if (stat != CUBLAS_STATUS_SUCCESS) {printf("data execution failed");cublasDestroy(handle);exit(-1);}cublasDestroy(handle);Error:    cudaFree(dev_c);    cudaFree(dev_a);    cudaFree(dev_b);        return cudaStatus;}

思路2：写两个内核函数，1个调用另外一个（待求证）。

__device__ void sum(const float *c, float* sum){}__global__ void calculate(const int *a, const int *b, float *c, float *sum){//calculate....int i = threadIdx.x;c[i] = a[i] + b[i];__syncthreads();sum(c, sum)}

思路3：写两个内核函数，顺序执行，即自己动手实现sum函数。

__global__ void calculate(const int *a, const int *b, float *c){//calculate....int i = threadIdx.x;c[i] = a[i] + b[i];}__global__ void sum(const float *c, float* sum){}

推荐思路1，因为尽量调用库函数（不自己手写），开发效率会提高。