將錯誤地從Fortran複製到cuda c程序的變量

Question

我一直在嘗試編寫一個程序，該程序使用GPU來使用高斯正交數值積分來計算積分。我一直在試圖弄清楚爲什麼這個程序不能正常工作。我想我把它固定在一個事實上，即在函數調用d_one中傳遞的參數沒有被正確地複製到cuda c代碼中。我不知道爲什麼會發生這種情況。我花了很多時間試圖弄清楚，但是我無法得到它。

這裏有兩個方案：

Fortran程序：

implicit real*8(a-h,o-z) 
    parameter (nlinx = 22) ! Total number of mesh regions 
    dimension sx(3*nlinx),swx(3*nlinx) 

    xa = 0.d0 
    xb = 5.d0 
    ! In the following "nptx" is the total number of integration 
    ! points. So, it is (nlinx * 3) 
    call meshwt1(xa,xb,nlinx,ntan,sx,swx,nptx) 

    ans0 = 0.d0 

    CAll d_one(sx, swx, nptx, ans0) 

    print *, ans0 

    stop 

    end 

SUBROUTINE MESHWT1(A,B,N,NT,X,W,NTOT) 
    implicit real*8(a-h,o-z) 
    !3*N LINEAR POINTS FOR A TO B 
    !NT=0 OR 1, 3*NT TAN PTS FOR B TO INFINITY 
    !NTOT= 3*(N+NT) 
    DIMENSION X(*),W(*),G(3),GW(3) 
    G(1) = -0.7745966 
    G(2) = 0.0000000 
    G(3) = -G(1) 
    GW(2) = 0.8888888 
    GW(1) = 0.5555555 
    GW(3) = GW(1) 
    Y = N 
    DX = (B - A)/Y 
    K = 0 
    XA = A - DX 
    XB = A 
    DO 2 I = 1, N 
    XA = XA + DX 
    XB = XB + DX 
    DO 2 J = 1, 3 
    K = K + 1 
    X(K) = 0.5 * (XA + XB) + 0.5 * (XB - XA) * G(J) 
2 W(K) = 0.5 * (XB - XA) * GW(J) 
    NTOT = K 
    IF(NT .EQ. 1) GO TO 3 
    GO TO 5 
3 NTOT = K + 3 
    DO 4 J = 1, 3 
    K = K + 1 
    Y = (1.0 + G(J)) * 3.14159 * 0.25 
    X(K) = XB + DTAN(Y) 
4 W(K) = GW(J) * 3.14159 * 0.25/(DCOS(Y)) ** 2 
5 CONTINUE 
    RETURN 
    END 

的CUDA程序：

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <cuda.h> 
#include <cuda_runtime.h> 
__global__ void loop_d(float *a, float *b, int N, float *ans) 
{ 
    __shared__ float temp[66]; 
    int idx = threadIdx.x; 
    if (idx < 66) 
    { 
       temp[idx] = a[idx] * b[idx]; 
    } 
    __syncthreads(); 
    if (0 == idx) 
    { 
      float sum = 0.0; 
      for (int i=0; i < 66; i++) 
      { 
        sum += temp[i]; 
      } 
      *ans = sum; 
    } 
} 
// The following function is called from the Fortran program 
extern "C" void d_one_(float *a, float *b, int *Np, float *ans) 
{ 
    float *a_d, *b_d, *ans_d; // Declaring GPU Copies of the parameters passed 
    int blocks = 1; // Number of blocks used 
    int N = *Np; // Number of threads is determined by the parameter nptx passed from the Fortran program 

    // Allocating GPU memory 
    cudaMalloc((void **)&a_d, sizeof(float) * N); 
    cudaMalloc((void **)&b_d, sizeof(float) * N); 
    cudaMalloc((void **)&ans_d, sizeof(float)); 
    // Copying information from CPU to GPU 
    cudaMemcpy(a_d, a, sizeof(float) * N, cudaMemcpyHostToDevice); 
    cudaMemcpy(b_d, b, sizeof(float) * N, cudaMemcpyHostToDevice); 
    cudaMemcpy(ans_d, ans, sizeof(float), cudaMemcpyHostToDevice); 
    // Calling the function on the GPU 
    loop_d<<< blocks, N >>>(a_d, b_d, N, ans_d); 
    cudaMemcpy(a, a_d, sizeof(float) * N, cudaMemcpyDeviceToHost); 
    cudaMemcpy(b, b_d, sizeof(float) * N, cudaMemcpyDeviceToHost); 
    cudaMemcpy(ans, ans_d, sizeof(float), cudaMemcpyDeviceToHost); 

    // Freeing GPU memory 
    cudaFree(a_d); 
    cudaFree(b_d); 
    cudaFree(ans_d); 
    return; 
} 

程序的輸出應該是12.49999。我得到了-314的答案。感謝您提供的任何輸入！

Answer 1

決定是要使用單精度浮點還是雙精度浮點變量。

目前，您正在使用Fortran方面的雙精度real*8和C（++）方面的單精度float。

一起使用real*4和float或real*8和double。