CUDA中的向量加法

时间:2018-12-01 08:36:20

标签: cuda

我正在尝试学习cuda编程,当我尝试从本教程here执行向量加法程序时,它没有按预期运行,没有引发任何错误,但是答案不正确。

#include<stdio.h> 
#include<stdlib.h> 

#define N 20 

__global__ void add(int *a, int *b, int *c){ 
        c[blockIdx.x] = a[blockIdx.x] + b[blockIdx.x]; 
} 
int main(void) { 
        int *a, *b, *c; // host copies of a, b, c 
        int *d_a, *d_b, *d_c; // device copies of a, b, c 
        int size = N * sizeof(int), i; 
        // Alloc space for device copies of a, b, c 
        cudaMalloc((void **)&d_a, size); 
        cudaMalloc((void **)&d_b, size); 
        cudaMalloc((void **)&d_c, size); 
        // Alloc space for host copies of a, b, c and setup input values 
        a = (int *)malloc(size);  
        b = (int *)malloc(size); 
        c = (int *)malloc(size); 
        for(i = 0; i< N ; i++){ 
                a[i] = rand() % 100; 
                b[i] = rand() % 50; 
        } 
        // Copy inputs to device 
        cudaMemcpy(d_a, a, size, cudaMemcpyHostToDevice); 
        cudaMemcpy(d_b, b, size, cudaMemcpyHostToDevice); 
        // Launch add() kernel on GPU with N blocks 
        add<<<N,1>>>(d_a, d_b, d_c); 

        cudaDeviceSynchronize(); 
        // Copy result back to host 
        cudaMemcpy(c, d_c, size, cudaMemcpyDeviceToHost); 
        // Cleanup 

        for(i = 0; i< N; i++) 
                printf("%d + %d = %d\n", a[i], b[i], c[i]); 
        free(a); free(b); free(c); 
        cudaFree(d_a); cudaFree(d_b); cudaFree(d_c); 
        return 0; 
} 

结果总是显示为-1,我不知道为什么,有人可以在这里帮助我吗?

1 个答案:

答案 0 :(得分:0)

我只是使用我的GPU来编译和运行程序,并且它按预期运行。您必须检查cudaMalloccudaMemCpy返回的错误代码,并确保它们没有返回任何错误。