Fortran 与 C++ 相比，如今 Fortran 在数值分析方面是否仍然具有优势 [关闭]

Fortran与C++在数值分析方面的比较

fortran
program linear_equation_solver
  implicit none
  integer, parameter :: n = 1000
  real :: A(n,n), b(n), x(n)
  integer :: i, j
  ! 初始化矩阵A和向量b
  do i = 1, n
    b(i) = 1.0
    do j = 1, n
      if (i == j) then
        A(i,j) = 2.0
      else
        A(i,j) = 1.0
      end if
    end do
  end do
  ! 解线性方程组
  call solve_linear_equations(A, b, x)
contains
  subroutine solve_linear_equations(A, b, x)
    implicit none
    real, dimension(:,:), intent(in) :: A
    real, dimension(:), intent(in) :: b
    real, dimension(:), intent(out) :: x
    ! 在这里实现线性方程组的求解算法
    ! ...
  end subroutine solve_linear_equations
end program linear_equation_solver

cpp
#include 
#include 
void solve_linear_equations(std::vector>& A, std::vector& b, std::vector& x)
{
  // 在这里实现线性方程组的求解算法
  // ...
}
int main()
{
  const int n = 1000;
  std::vector> A(n, std::vector(n));
  std::vector b(n);
  std::vector x(n);
  // 初始化矩阵A和向量b
  for (int i = 0; i < n; ++i)
  {
    b[i] = 1.0;
    for (int j = 0; j < n; ++j)
    {
      if (i == j)
      {
        A[i][j] = 2.0;
      }
      else
      {
        A[i][j] = 1.0;
      }
    }
  }
  // 解线性方程组
  solve_linear_equations(A, b, x);
  return 0;
}

fortran
program parallel_computation
  implicit none
  integer, parameter :: n = 1000000
  real :: A(n), B(n), C(n)
  integer :: i
  ! 初始化数组A和B
  do i = 1, n
    A(i) = 1.0
    B(i) = 2.0
  end do
  ! 并行计算
  !$OMP PARALLEL DO DEFAULT(NONE) SHARED(A, B, C) PRIVATE(i)
  do i = 1, n
    C(i) = A(i) + B(i)
  end do
  !$OMP END PARALLEL DO
end program parallel_computation

cpp
#include 
#include 
#include 
int main()
{
  const int n = 1000000;
  std::vector A(n);
  std::vector B(n);
  std::vector C(n);
  // 初始化数组A和B
  for (int i = 0; i < n; ++i)
  {
    A[i] = 1.0;
    B[i] = 2.0;
  }
  // 并行计算
#pragma omp parallel for
  for (int i = 0; i < n; ++i)
  {
    C[i] = A[i] + B[i];
  }
  return 0;
}