来自 thrust::reduce 的总和值不正确

Question

我一直在尝试实现一些需要在 thrust::device_ptr 上调用 reduce 的代码，但在处理大值时，结果与 CPU 的实现不一致。我必须处理大值。那么有没有办法解决：

我的代码：

    #include <cuda_runtime_api.h>
    #include <stdio.h>
    #include <thrust/host_vector.h>
    #include <thrust/device_vector.h>
    #include <iostream>
    #define   NZ  412//
    #define   NX  402//
    using namespace std;
    using real =double;
    
    
    void allocate_array_2d(real**& preal, const int dim1, const int dim2) {
        // Contiguous allocation of 2D arrays
    
        preal = new real * [dim1];
        preal[0] = new real[dim1 * dim2];
        for (int i = 1; i < dim1; i++) preal[i] = preal[i - 1] + dim2;
    
        for (int i = 0; i < dim1; i++) {
            for (int j = 0; j < dim2; j++) {
                preal[i][j] = 0;
            }
        }
    }
    #define cudaCheckError(code)                                             \
      {                                                                      \
        if ((code) != cudaSuccess) {                                         \
          fprintf(stderr, "Cuda failure %s:%d: '%s' \n", __FILE__, __LINE__, \
                  cudaGetErrorString(code));                                 \
        }                                                                    \
      }
    
    
    int main()
    
    {
        real** a;
        std::cout.precision(30);
        allocate_array_2d(a, NZ, NX);//input array
       
        for (int i = 0; i < NZ; i++) {
            for (int j = 0; j < NX; j++) {
                a[i][j] = 2.14748e+09;
              
              
            }
        }
    
            real* da;
            cudaCheckError(cudaMalloc(&da, NZ * NX  * sizeof(real)));
            cudaCheckError(cudaMemcpy(da,a[0], NZ * NX  * sizeof(real),cudaMemcpyHostToDevice));
    
            ///************************
            //CUDA KERNELS ARE HERE
            // REMOVED FOR CLEAR QUESTION
            ///*************************
      
            real sum1=0;
          
            thrust::device_ptr<real> dev_ptr = thrust::device_pointer_cast(da);
            sum1 = thrust::reduce(dev_ptr, dev_ptr+NZ*NX, 0, thrust::plus<real>());
            
            cout<<" \nsum gpu "<< sum1<<"\n";
    
            real sum2=0;
    
            ////////CPU PART DOING SAME THING//////
            for (int i = 0; i < NZ; i++) {
    
                for (int j = 0; j < NX; j++) {
                   sum2 += a[i][j];
                    
                }
            }
    
    
            cout<<"\nsum cpu "<< sum2<<"\n";
            if((sum2-sum1)<0.001)
            std::cout << "\nSUCESS "<< "\n";
            else
            std::cout << "\nFailure & by "<<sum2-sum1<< "\n";
       
    }

我使用的编译器是nvcc，我的显卡是nvidia 1650，计算能力7.5。

Answer 1

根据the documentation，推力期望求和的类型反映在init值中：

sum1 = thrust::reduce(dev_ptr, dev_ptr+NZ*NX, 0, thrust::plus<real>());
                                              ^

你的常量类型是整型。如果将其更改为双精度常量：

sum1 = thrust::reduce(dev_ptr, dev_ptr+NZ*NX, 0.0, thrust::plus<real>());

根据我的测试，您会得到 CPU 和 GPU 之间的匹配结果。（您也可以将常量转换为 real 类型：(real)0 并使用它，还有其他方法可以解决这个问题，例如放弃使用初始值和二进制操作。）

来自 thrust::reduce 的总和值不正确

Value of sum from thrust::reduce not correct

parallel-processing

cuda

hpc

nvidia

thrust