无法打印从 CUDA 内核返回的值

Question

我正在研究在同一节点交换 V1 和 V3 的变量。但是，我无法将值 16 和 31 初始化为数组。这可能是我犯的一个愚蠢的错误，但我已经花了一个小时来调试我的代码。它只在 printf 输出中打印每个数组的“0”。

谁能找出我代码中的错误？这是我的代码：

 #include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdio.h>
#include <cuda.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>


#define threads 1024  //define the number of thread to use
#define blocks 4 //define the number of blocks to use


//Kernal Function
__global__ void Initial(double* V1, double* V3, int NX, int NY)
{
    unsigned int idx = threadIdx.x + blockIdx.x * blockDim.x;

    if (idx <= NX * NY)
    {
        V1[idx] = 16;
        V3[idx] = 31;
    }

}


__global__ void Add2D(double* V1 , double* V3, int NX, int NY)
{
    unsigned int idx = threadIdx.x + blockIdx.x * blockDim.x;
    

    if (idx <= NX * NY)
    {
        double tmp = V1[idx] ;
        V1[idx] = V3[idx + NX];
        V3[idx + NX] = tmp;

    }
}


int main(void) {

    //number of nodes 
    int NX = 5;
    int NY = 5;
    int N = NY * NX;

    size_t bytes =  NX * NY * sizeof(double);  // define the memory size which needs to use in this application

    //declare V1 in host
    double* hos_V1 ;
    double* hos_V3 ;

    double* dev_V1 ;
    double* dev_V3 ;

//  hos_V1 = new double[N];     // allocate storage for VL array
//  hos_V3 = new double[N];

    hos_V1 = (double*)malloc(bytes);
    hos_V3 = (double*)malloc(bytes);

//  dev_V1 = new double[N];     // allocate storage for VL array
//  dev_V3 = new double[N];


    cudaMalloc((void**)&dev_V1, bytes);
    cudaMalloc((void**)&dev_V3, bytes);


    Initial <<< blocks , threads >>> (dev_V1, dev_V3, NX, NY);

    cudaMemcpy(hos_V1, dev_V1, bytes, cudaMemcpyDeviceToHost);
    cudaMemcpy(hos_V3, dev_V3, bytes, cudaMemcpyDeviceToHost);

    cudaThreadSynchronize();

    for (int x = 0; x <= NX * NY; x++)
    {

        printf("V1[%d] = %f  \n", x, &hos_V1[x]);
        printf("V3[%d] = %f  \n", x, &hos_V3[x]);

    }
    printf("-----------------------------\n");

    Add2D <<< blocks, threads >> > (dev_V1, dev_V3, NX, NY);

    cudaThreadSynchronize(); //Sync CPU and GPU to start the timer  
    cudaMemcpy(hos_V1, dev_V1, bytes, cudaMemcpyDeviceToHost);
    cudaMemcpy(hos_V3, dev_V3, bytes, cudaMemcpyDeviceToHost);

    for (int x = 0; x <= NX*NY; x++)
    {
        
            printf("V1[%d] = %f  \n", x, &hos_V1[x]);
            printf("V3[%d] = %f  \n", x, &hos_V3[x]);
        
    }

    // free the memory allocated on the GPU
    cudaFree(dev_V1);
    cudaFree(dev_V3);


    return 0;
}

Answer 1

您正在打印 &hos_V1[x] 和 &hos_V3[x]，而不是 hos_V1[x] 和 hos_V3[x]。我确定您想实际打印数组的内容。

for (int x = 0; x <= NX * NY; x++)
{

    printf("V1[%d] = %f  \n", x, hos_V1[x]);
    printf("V3[%d] = %f  \n", x, hos_V3[x]);

}

在编译时启用警告可能对您有用。 NVCC（或 gcc，因为它是主机代码）向我建议您可能正在打印错误的东西。 edit:as 其他人指出，您提供的代码还有更多“系统性”问题。

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