(CUDA C)为什么不打印出从设备存储器复制的值? [英] (CUDA C) Why is it not printing out the value copied from device memory?

问题描述

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，显示了如何将两个整数相加.代码如下:

I'm learning CUDA right now through the training slides provided by NVIDIA. They have a sample program that shows how you could add two integers. The code is below:

#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，输出为: 对于a = 2，b = 7，我们得到a + b = -3"

But when I run the program, the output is: "For a = 2, b = 7, we get a + b = -3"

表示c的值不变！

我在做什么错了?

推荐答案

您的代码正确地将c的值打印为9.您需要弄清正在运行此代码的环境.

Your code is correctly printing the value of c as 9. You need to clarify on the environment you are running this code.

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