Question

我试图理解指针在CUDA中的功能。

下面给出的是一个简单的程序，它为设备中分配的变量赋予特定值。（我使用带有NVIDIA Quadro K2000显卡的CUDA工具包8.0）

当我使用printf（）在内核中打印值时，它显示错误的值。

但是，当我从主机功能执行cudaMemcopy然后从主机功能打印时，它会显示正确的值......

//CODE...
#include <stdio.h>
#include <stdlib.h>
#include<cuda.h>
#include<cuda_runtime.h>
#define N 3

__global__ void tempker(int *jk,int value)
{
    (*jk) = value*2;
    printf("displayed from inside the kernel :\nvalue of jk = %d\nvalue of *jk = %d\n",jk,*jk);

}
int tempfunc(int *kp)
{
    int *jk = NULL,*lm=NULL;
    lm = (int *)(malloc(sizeof(int)));
    *lm = 150;
    cudaError_t err = cudaSuccess;
    cudaMalloc((void**)&jk, sizeof(int));
    printf("jk pointer after cudaMalloc: displayed from host = %d\n",jk);
    tempker<<<1,1>>>(jk,150);
    err = cudaGetLastError();//brief Returns the last error from a runtime call
    cudaDeviceSynchronize();

    err = cudaMemcpy(lm, jk, (sizeof(int)), cudaMemcpyDeviceToHost);
    if (err != cudaSuccess)
    {
        fprintf(stderr, "Failed to copy jk from device to host (error code %s)!\n", cudaGetErrorString(err));
        exit(EXIT_FAILURE);
    }
    printf("Displayed in host function after memcopy: value of *lm = *jk = %d\n",*lm);
    cudaFree(jk);
    err = cudaMalloc((void**)&kp, sizeof(int));
    if (err != cudaSuccess)
    {
        fprintf(stderr, "Failed to allocate device kp (error code %s)!\n", cudaGetErrorString(err));
        exit(EXIT_FAILURE);
    }
    tempker<<<1,1>>>(kp,(N*N*N));
    err = cudaGetLastError();//brief Returns the last error from a runtime call
    cudaDeviceSynchronize();

    err = cudaMemcpy(lm, kp, (sizeof(int)), cudaMemcpyDeviceToHost);
    if (err != cudaSuccess)
    {
        fprintf(stderr, "Failed to copy kp from device to host (error code %s)!\n", cudaGetErrorString(err));
        exit(EXIT_FAILURE);
    }
    printf("Displayed in host function after memcopy: value of *lm = *kp = %d\n",*lm);
    cudaFree(kp);
    free(lm);
    return 100;
}

int main(){

    int *kp = NULL;
    printf("tempfunc(): return value = %d\n",tempfunc(kp));
    return 0;
}

输出：

jk pointer after cudaMalloc: displayed from host = 13238272
displayed from inside the kernel :
value of jk = 13238272
value of *jk = 9
Displayed in host function after memcopy: value of *lm = *jk = 300
displayed from inside the kernel :
value of jk = 13238272
value of *jk = 9
Displayed in host function after memcopy: value of *lm = *kp = 54
tempfunc(): return value = 100

问题是： 是否可以打印内核中设备中分配的变量的值？

Answer 1

使用“％d”显示整数值被认为是正确的。使用“％d”显示地址可能会导致不可预测的行为，具体取决于您的编译器。

只需使用“％p”即可。因为它需要一个void指针，所以明确地转换你的指针。

所以在你的内核中：

printf("value = %d, address = %p\n",*jk,(void *)jk);

如果您进一步了解printf，可以获得更多信息。

无法在CUDA中的内核中打印设备变量值

1 个答案: