Question

我真的不明白这两个程序之间的区别。 Fortran可以正常工作并提供正确的结果

81  88 221 335 778 595 1559 1759

虽然C ++程序失败并给出了这个结果：

81 181 221 335 778 595 1559 1759

正如您所看到的，第二个元素是不同的，正确的值是使用Fortran代码获得的值：

 MODULE MSR
 IMPLICIT NONE

CONTAINS
     subroutine amuxms (n, x, y, a,ja)
      real*8  x(*), y(*), a(*)
      integer n, ja(*)

      integer i, k

        do 10 i=1, n
        y(i) = a(i)*x(i)
 10     continue
      do 100 i = 1,n

         do 99 k=ja(i), ja(i+1)-1

            y(i) = y(i) + a(k) *x(ja(k))
 99      continue
 100  continue

      return

      end

END MODULE

  PROGRAM MSRtest
        USE MSR
        IMPLICIT NONE
         INTEGER :: i
         REAL(KIND(0.D0)), DIMENSION(8) :: y, x= (/3,4,0,1,6,8,1,19/)

         REAL(KIND(0.D0)), DIMENSION(19) :: AA = (/11,22,33,44,55,66,77,88,0,12,31,32,41,42,43,56,67,78,87/) 
         INTEGER , DIMENSION(19)         :: JA = (/10,11,11,13,16,17,18,19,20,2,1,2,1,2,3,6,7,8,7/)




         CALL amuxms(8,x,y,aa,ja)

         WRITE(6,FMT='(8F10.2)') (y(I), I=1,8)    

 END PROGRAM

为什么这个C ++程序会产生不同的结果？

  # include <iomanip>
  # include <iostream>

std::vector<double> dot( int size, std::vector<double> x, 
                         std::vector<double> aa, std::vector<int> ja)
{     

      std::vector<double> y(x.size());

      for(auto i = 0; i < size ; i++ )
            y.at(i) = aa.at(i) * x.at(i);

      std::vector<double> jja(ja.size());

      for(auto i=0 ; i < size ; i++ )
      {

         auto k = ja.at(i)-1;

         do
         {
            y.at(i) += aa.at(k) * x.at(ja.at(k)-1) ;
            k++;
         }
         while(k < ja.at(i+1)-1) ;

      }
      return y;
}


int main()
{
    std::vector<double> x =  {3,4,0,1,6,8,1,19};


    std::vector<double> aa = {11,22,33,44,55,66,77,88,0,12,31,32,41,42,43,56,67,78,87}; 
    std::vector<int>    ja =  {10,11,11,13,16,17,18,19,20,2,1,2,1,2,3,6,7,8,7};//{6,7,7,8,10,3,1,1,3};

    std::vector<double> y = dot(x.size(), x , aa , ja);

    for(auto& i : y)
         std::cout << i << ' ' ;   
    std::cout << std::endl;  
}

奇怪的是，如果我给这个载体：

AA = {1.01,4.07,6.08,9.9,0.,2.34,3.12,1.06,2.2};
JA = {6,7,7,8,10,3,1,1,3};
X  = {0.,1.3,4.2,0.8}

两个程序都给出了正确的结果：

9.828 5.291 25.536 17.16

Answer 1

改为：

   # include <iomanip>
    # include <iostream>

    std::vector<double> dot( int size, std::vector<double> x, 
                             std::vector<double> aa, std::vector<int> ja)
    {     

          std::vector<double> y(x.size());

          for(auto i = 0; i < size ; i++ )
                y.at(i) = aa.at(i) * x.at(i);

          std::vector<double> jja(ja.size());
    //      for(auto j=0 ; j< jja.size() ;  j++)
      //          jja.at(j) = ja.at(j)-1 ;

          for(auto i=0 ; i < size ; i++ )
          {
             //for(auto k=ja.at(i) ; k< ja.at(i+1)-1 ; k++ )
             //auto k = ja.at(i)-1;
             //std::cout << ja.at(i) <<"  " << i << std::endl ;
            // do
            for(auto k=ja.at(i)-1 ; k< ja.at(i+1)-1 ; k++ )
             {
                y.at(i) += aa.at(k) * x.at(ja.at(k)-1) ;
             }
            // while(k < ja.at(i+1)-1) ;

          }
          return y;
    }

int main()
{
    std::vector<double> x =  {3,4,0,1,6,8,1,19};


    std::vector<double> aa = {11,22,33,44,55,66,77,88,0,12,31,32,41,42,43,56,67,78,87}; 
    std::vector<int>    ja =  {10,11,11,13,16,17,18,19,20,2,1,2,1,2,3,6,7,8,7};//{6,7,7,8,10,3,1,1,3};

    std::vector<double> y = dot(x.size(), x , aa , ja);

    for(auto& i : y)
         std::cout << i << ' ' ;   
    std::cout << std::endl;  
}

给出：81 88 221 335 778 595 1559 1759 使用vector作为数组索引时请注意！

类似的程序有不同的结果

1 个答案: