c - (CUDA C) 为什么它不打印从设备内存复制的值？

Question

我现在正在通过 NVIDIA 提供的培训幻灯片学习 CUDA。他们有一个示例程序，展示了如何将两个整数相加。代码如下:

#include <stdio.h>

__global__ void add(int *a, int *b, int *c) {
    *c = *a+*b;
}

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
    size_t size = sizeof(int);

    //Allocate space for device copies of a, b, c
    cudaMalloc((void**)&d_a, size);
    cudaMalloc((void**)&d_b, size);
    cudaMalloc((void**)&d_c, size);

    //Setup input values
    a = 2;
    b = 7;
    c = -3;

    //Copy inputs to device
    cudaMemcpy(d_a, &a, size, cudaMemcpyHostToDevice);
    cudaMemcpy(d_b, &b, size, cudaMemcpyHostToDevice);

    //Launch add() kernel on GPU
    add<<<1,1>>>(d_a, d_b, d_c);

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

    //Cleanup
    cudaFree(d_a); cudaFree(d_b); cudaFree(d_c);

    printf("For a = %d, b = %d, we get a + b = %d\n", a, b, c);

    return 0;
}

但是当我运行程序时，输出是:“对于 a = 2，b = 7，我们得到 a + b = -3”

意味着c的值没有改变!

我做错了什么？

Answer 1

您的代码正确地将 c 的值打印为 9。您需要澄清运行此代码的环境。