问题描述
我正在尝试计算curand_uniform()返回1.0的次数。但是我似乎无法获得以下代码为我工作:
#include <stdio.h>
#include <stdlib.h>
#include <thrust/device_vector.h>
#include <cuda.h>
#include <cuda_runtime.h>
#include <curand_kernel.h>
using namespace std;
__global__
void counts(int length,int *sum,curandStatePhilox4_32_10_t* state) {
int tempsum = int(0);
int i = blockIdx.x * blockDim.x + threadIdx.x;
curandStatePhilox4_32_10_t localState = state[i];
for(; i < length; i += blockDim.x * gridDim.x) {
double thisnum = curand_uniform( &localState );
if ( thisnum == 1.0 ){
tempsum += 1;
}
}
atomicAdd(sum,tempsum);
}
__global__
void curand_setup(curandStatePhilox4_32_10_t *state,long seed) {
int id = threadIdx.x + blockIdx.x * blockDim.x;
curand_init(seed,id,&state[id]);
}
int main(int argc,char *argv[]) {
const int N = 1e5;
int* count_h = 0;
int* count_d;
cudaMalloc(&count_d,sizeof(int) );
cudaMemcpy(count_d,count_h,sizeof(int),cudaMemcpyHostToDevice);
int threads_per_block = 64;
int Nblocks = 32*6;
thrust::device_vector<curandStatePhilox4_32_10_t> d_state(Nblocks*threads_per_block);
curand_setup<<<Nblocks,threads_per_block>>>(d_state.data().get(),time(0));
counts<<<Nblocks,threads_per_block>>>(N,count_d,d_state.data().get());
cudaMemcpy(count_h,cudaMemcpyDeviceToHost);
cout << count_h << endl;
cudaFree(count_d);
free(count_h);
}
我遇到终端错误(在 linux):
terminate called after throwing an instance of 'thrust::system::system_error'
what(): parallel_for failed: cudaErrorInvalidValue: invalid argument
Aborted (core dumped)
我正在这样编译:
nvcc -Xcompiler "-fopenmp" -o test uniform_one_hit_count.cu
我不明白此错误消息。
解决方法
此行:
thrust::device_vector<curandStatePhilox4_32_10_t> d_state(Nblocks*threads_per_block);
正在初始化设备上的新向量。当推力这样做时,它将调用使用中的对象的构造函数,在本例中为curandStatePhilox4_32_10
,该结构的定义在/usr/local/cuda/include/curand_philox4x32_x.h
中(无论如何,在Linux上)。不幸的是,该结构定义没有提供任何用__device__
装饰的构造函数,这给推力带来了麻烦。
一个简单的解决方法是在主机上组装向量并将其复制到设备:
thrust::host_vector<curandStatePhilox4_32_10_t> h_state(Nblocks*threads_per_block);
thrust::device_vector<curandStatePhilox4_32_10_t> d_state = h_state;
或者,只需使用cudaMalloc分配空间:
curandStatePhilox4_32_10_t *d_state;
cudaMalloc(&d_state,(Nblocks*threads_per_block)*sizeof(d_state[0]));
您也至少还有一个其他问题。实际上,这并未为指针应指向的内容提供适当的存储分配:
int* count_h = 0;
之后,您应该执行以下操作:
count_h = (int *)malloc(sizeof(int));
memset(count_h,sizeof(int));
在您的打印输出行上,您最有可能希望这样做:
cout << count_h[0] << endl;
解决count_h
问题的另一种方法是从以下开始:
int count_h = 0;
,这将需要对代码(cudaMemcpy
操作)进行不同的更改。