CUDA编程（一）：实现两个数组相加

前言

 由于最近忙着秋招，本系列博客最近只记录下自己学CUDA的例子，不会做出注释，后期有空会补上基础知识和代码详解。

---

#include <stdio.h>

const double a = 1.23;
const double b = 2.34;

void __global__ add(const double *x, const double *y, const double *z,const int N );

void __global__ add(const double *x, const double *y, double *z, const int N)
{
    
    
    const int tid = blockDim.x * blockIdx.x + threadIdx.x;
    if(tid < N)
    {
    
    
    z[tid] = x[tid] + y[tid];
    }
}

int main()
{
    
    
    const int N = 1000;
    const int M = sizeof(double) * N;

    double *ha = (double *)malloc(M);
    double *hb = (double *)malloc(M);
    double *hc = (double *)malloc(M);
    // assignment

    for(int i=0; i < N; ++i)
    {
    
    
        ha[i] = a;
        hb[i] = b;
    }

    //
    double *da, *db, *dc;

    cudaMalloc((void**)&da, M);
    cudaMalloc((void**)&db, M);
    cudaMalloc((void**)&dc, M);

    cudaMemcpy(da,ha,M,cudaMemcpyHostToDevice);
    cudaMemcpy(db,hb,M,cudaMemcpyHostToDevice);

    // kernel fun

    const int block_size = 128;
    const int grid_size = (N + block_size -1)/ block_size;
    add<<<grid_size,block_size>>>(da,db,dc,N);

    cudaMemcpy(hc,dc,M,cudaMemcpyDeviceToHost);

    free(ha);
    free(hb);
    free(hc);
    cudaFree(da);
    cudaFree(db);
    cudaFree(dc);

    return 0;

}

  编译指令

nvcc -arch=sm_75 add.cu -o add