Cuda 通用点积
我对C和CUDA都是新手,我在写点积函数,但是它没有给我正确的结果。有没有善良的灵魂能帮我看看 我还有两个问题,Cuda 通用点积,cuda,dot-product,Cuda,Dot Product,我对C和CUDA都是新手,我在写点积函数,但是它没有给我正确的结果。有没有善良的灵魂能帮我看看 我还有两个问题, 为什么dot()不能正常工作,以及 在第57行,为什么是product[threadIdx.x]而不是product[index]?我不能写吗 产品[指数]=a[指数]*b[指数]; ... 如果(索引==0){…} 然后用第零个线程对每个元素求和 非常感谢 设备查询: Device 0: "GeForce GTX 570" CUDA Driver Version / Run
Device 0: "GeForce GTX 570"
CUDA Driver Version / Runtime Version 6.0 / 5.5
CUDA Capability Major/Minor version number: 2.0
Makefile:nvcc-arch=sm_20 cuda_test.cu-o cuda_test
在cuda_test.cu中:
#include <stdio.h> // printf, scanf, NULL etc.
#include <stdlib.h> // malloc, free, rand etc.
#define N (3) //Number of threads we are using (also, length of array declared in main)
#define THREADS_PER_BLOCK (1) //Threads per block we are using
#define N_BLOCKS (N/THREADS_PER_BLOCK)
/* Function to generate a random integer between 1-10 */
void random_ints (int *a, int n)
{
int i;
srand(time(NULL)); //Seed rand() with current time
for(i=0; i<n; i++)
{
a[i] = rand()%10 + 1;
}
return;
}
/* Kernel that adds two integers a & b, stores result in c */
__global__ void add(int *a, int *b, int *c) {
//global indicates function that runs on
//device (GPU) and is called from host (CPU) code
int index = threadIdx.x + blockIdx.x * blockDim.x;
//threadIdx.x : thread index
//blockIdx.x : block index
//blockDim.x : threads per block
//hence index is a thread counter across all blocks
c[index] = a[index] + b[index];
//note that pointers are used for variables
//add() runs on device, so they must point to device memory
//need to allocate memory on GPU
}
/* Kernel for dot product */
__global__ void dot(int *a, int *b, int *c)
{
__shared__ int product[THREADS_PER_BLOCK]; //All threads in a block must be able
//to access this array
int index = threadIdx.x + blockIdx.x * blockDim.x; //index
product[threadIdx.x] = a[index] * b[index]; //result of elementwise
//multiplication goes into product
//Make sure every thread has finished
__syncthreads();
//Sum the elements serially to obtain dot product
if( 0 == threadIdx.x ) //Pick one thread to sum, otherwise all will execute
{
int sum = 0;
for(int j=0; j < THREADS_PER_BLOCK; j++) sum += product[j];
//Done!
atomicAdd(c,sum);
}
}
int main(void)
{
int *a, *b, *c, *dotProduct; //host copies of a,b,c etc
int *d_a, *d_b, *d_c, *d_dotProduct; //device copies of a,b,c etc
int size = N * sizeof(int); //size of memory that needs to be allocated
int i=0; //iterator
//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 = (int *)malloc(size); random_ints(a,N);
b = (int *)malloc(size); random_ints(b,N);
c = (int *)malloc(size);
//Copy inputs to device
cudaMemcpy(d_a, a, size, cudaMemcpyHostToDevice);
cudaMemcpy(d_b, b, size, cudaMemcpyHostToDevice);
//Launch add() kernel on GPU
add<<<N_BLOCKS,THREADS_PER_BLOCK>>>(d_a, d_b, d_c);
// triple angle brackets mark call from host to device
// this is also known as a kernel launch
// N/THREADS_PER_BLOCK = NO. OF BLOCKS
//Copy result back to host
cudaMemcpy(c, d_c, size, cudaMemcpyDeviceToHost);
//Output results
printf("a = {");
for (i=0; i<N; i++) printf(" %d",a[i]);
printf(" }\n");
printf("b = {");
for (i=0; i<N; i++) printf(" %d",b[i]);
printf(" }\n");
printf("c = {");
for (i=0; i<N; i++) printf(" %d",c[i]);
printf(" }\n");
//Calculate dot product of a & b
dotProduct = (int *)malloc(sizeof(int)); //Allocate host memory to dotProduct
*dotProduct = 0; //initialise to zero
cudaMalloc((void **)&d_dotProduct, sizeof(int)); //Allocate device memory to d_dotProduct
dot<<<N_BLOCKS,THREADS_PER_BLOCK>>>(d_a, d_b, d_dotProduct); //Perform calculation
cudaMemcpy(dotProduct, d_dotProduct, sizeof(int), cudaMemcpyDeviceToHost); //Copy result into dotProduct
printf("\ndot(a,b) = %d\n", *dotProduct); //Output result
//Cleanup
free(a); free(b); free(c); free(dotProduct);
cudaFree(d_a); cudaFree(d_b); cudaFree(d_c); cudaFree(d_dotProduct);
return 0;
} //End of main
#包括//printf、scanf、NULL等。
#包括//malloc、free、rand等。
#define N(3)//我们正在使用的线程数(还有main中声明的数组长度)
#定义每个块的线程数(1)//我们正在使用的每个块的线程数
#定义N个块(每个块有N个线程)
/*函数生成1-10之间的随机整数*/
无效随机整数(整数*a,整数n)
{
int i;
srand(time(NULL));//使用当前时间设置rand()
对于(i=0;i,正如Talonmes所说,请将其设置为其他人可以运行您的代码。嵌入行号是没有帮助的
在没有其他信息的情况下,最好的猜测是您没有将d_dotProduct
初始化为零。您可以使用cudaMemset()
-如果您需要不同的初始值,那么您可以cudaMemcpy()
从主机获取初始值或启动单独的内核进行初始化,但在这种情况下,cudaMemset()
(相当于主机上的memset()
)就足够了
也可能是N个块*每个块的线程数
不等于size
至于你的第二个问题,product
是一个大小为THREADS\u per\u block
的每块数组,如果你使用product[index]
访问它,你将无法访问它。问题已解决!需要在对“product”数组的各个元素求和之前设置“*c=0”
/* Kernel for dot product */
__global__ void dot(int *a, int *b, int *c)
{
__shared__ int product[THREADS_PER_BLOCK]; //All threads in a block must be able
//to access this array
int index = threadIdx.x + blockIdx.x * blockDim.x; //index
product[threadIdx.x] = a[index] * b[index]; //result of elementwise
//multiplication goes into product
if(index==0) *c = 0; //Ask one thread to set c to zero.
//Make sure every thread has finished
__syncthreads();
//Sum the elements serially to obtain dot product
if( 0 == threadIdx.x ) //Every block to do c += sum
{
int sum = 0;
for(int j=0; j < THREADS_PER_BLOCK; j++) sum += product[j];
//Done!
atomicAdd(c,sum);
}
}
/*用于dot产品的内核*/
__全局无效点(int*a,int*b,int*c)
{
__shared_uuuint product[THREADS_PER_BLOCK];//块中的所有线程都必须能够
//要访问此阵列
int index=threadIdx.x+blockIdx.x*blockDim.x;//索引
product[threadIdx.x]=a[index]*b[index];//elementwise的结果
//乘法变成乘积
if(index==0)*c=0;//要求一个线程将c设置为零。
//确保每根线都已完成
__同步线程();
//将元素按顺序求和以获得点积
if(0==threadIdx.x)//要执行的每个块c+=sum
{
整数和=0;
对于(int j=0;j
请发布没有嵌入行号的代码,其他人可以根据自己的意愿编译和运行这些代码。此外,请解释什么“工作不正常”意思正是。嗨,Talonmes,对不起!我已经清理了它,现在应该可以编译了。很抱歉早些时候浪费了你的时间!嗨,tom,谢谢你回答问题2!我已经清理了代码并完整地包含了它。很抱歉早些时候浪费了你的时间!这是错误的,只有当你启动一个threadblock时它才会工作。如果你启动多个threadblock,那么每个块的线程0会将最终结果重置为零。仅在块0中执行此操作也是不正确的,因为编程模型允许块以任何顺序执行。正确的解决方案是在启动内核之前将d_dotProduct设置为零。现在呢?我添加了if(index==0)*c=0行。当d_dotProduct仅存在于GPU内存中时,如何从主机代码将其设置为零?您肯定必须从内核中进行设置?另一方面,非常感谢您的帮助,tom。正如我上面所说,您不能只在块零中进行设置,因为块可以以任何顺序执行,例如块1可以在块之前写入其结果0将结果(覆盖块1的结果)重置为零。正确的方法是在启动内核之前将初始值设置为零-最明显的方法是使用cudaMemset()
,但也可以使用cudaMemcpy()
复制初始值(在本例中为零)来自主持人。我将在我的答案中添加how部分。