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Oop 为不同的(曲线生成)类编写一个可移植的(根查找器)解算器类_Oop_Fortran_Portable Class Library_Fortran2003 - Fatal编程技术网

Oop 为不同的(曲线生成)类编写一个可移植的(根查找器)解算器类

Oop 为不同的(曲线生成)类编写一个可移植的(根查找器)解算器类,oop,fortran,portable-class-library,fortran2003,Oop,Fortran,Portable Class Library,Fortran2003,我非常渴望实现FORTRAN 2003(F2003)的面向对象编程(OOP)功能。我的问题更多的是关于程序的设计。假设我有一个解算器,就像一个函数的根查找器f(x)=0;在最简单的FORTRAN格式中,它将得到 function solver(f,a,b) result(root) [some definition of variables] [some iterative procedures] root = ... end function 在以前版本的FORTRA

我非常渴望实现FORTRAN 2003(F2003)的面向对象编程(OOP)功能。我的问题更多的是关于程序的设计。假设我有一个解算器,就像一个函数的根查找器f(x)=0;在最简单的FORTRAN格式中,它将得到

function solver(f,a,b) result(root)
    [some definition of variables]
    [some iterative procedures]
    root = ...
end function
在以前版本的FORTRAN(如f95)中,为了获得可移植性,代码单独编译,并将外部函数传递给解算器。 现在从F2003面向对象的角度来看,假设在一般情况下,我们有一个用于求解器的类

我们的曲线还有一节课

type curve_t
    [some variable definition]
contains
    [some initialization procedures]
    procedure :: func => function_curve_t    ! y=this%func(x)
    procedure :: plot => plot_curve
end type

而且会有更多不同的曲线类(类型)。现在,我如何以编译解算器类(不知道曲线类/类型)的方式连接这两个概念,并且无论何时编写新的不同曲线类(如二阶多项式曲线、三阶多项式曲线、对数曲线、经验曲线等),我都能够实现它而无需更改。我的意思是,在最后,我以某种方式得到了曲线的根。

这里是一个如何使用F2003 OOP实现这个想法的示例。我将从构建到共享库中的模块开始:

module solver
  implicit none

  type, abstract :: curve_t
   contains
     procedure(func_f), pass(this), deferred :: f
  end type curve_t

  type :: solver_t
     class(curve_t), pointer :: curve
   contains
     procedure, pass :: solve => solve_root_bisect_method
  end type solver_t

  abstract interface
     function func_f(this, x)
       import curve_t
       class(curve_t) :: this
       real, intent(in) :: x
       real :: func_f
     end function func_f
  end interface

contains

  function solve_root_bisect_method(this, a_start, b_start) result(root)
    implicit none
    class(solver_t) :: this
    real, intent(in) :: a_start, b_start
    real :: root, c, eps, a, b
    integer :: i, imax
    imax = 100
    eps = 1e-5
    a = a_start
    b = b_start

    do i=1, imax
       c = (a+b)/2.
       if ( (this%curve%f(c) == 0) .or. ((b-a)/2. < eps)) then
          root = c
          return
       end if
       if (sign(1.,this%curve%f(c)) == sign(1.,this%curve%f(a))) then
          a = c
       else
          b = c
       end if
    end do
    ! solution did not converge, produce error
    root = -999
  end function solve_root_bisect_method
end module solver
这将产生
solver.so
solver.mod
。我做了这个额外的步骤来演示可移植性和编译,而不需要了解任何曲线

现在,我们可以假装是第三方,希望使用这个方便的库来查找任意曲线的根。首先,我们可以定义自己的模块来扩展曲线,并提供一些函数

module curves
  use solver
  implicit none

  type, extends(curve_t) :: linear_curve
     real :: m, b
   contains
     procedure, pass(this) :: f => f_linear
  end type linear_curve

  type, extends(curve_t) :: polynomial_curve
     real :: a, b, c
   contains
     procedure, pass(this) :: f => f_polynomial
  end type polynomial_curve

contains

  real function f_linear(this, x)
    use solver
    implicit none
    class(linear_curve) :: this
    real, intent(in) :: x
    f_linear = this%m * x + this%b
  end function f_linear

  real function f_polynomial(this, x)
    use solver
    implicit none
    class(polynomial_curve) :: this
    real, intent(in) :: x
    f_polynomial = this%a*x*x + this%b*x + this%c
  end function f_polynomial
end module curves
这定义了包含其参数的线性曲线和多项式曲线的类型,以及一个用于计算
y
的函数,作为给定这些参数的
x
的函数。因为我们从
curve\u t
派生,并且符合
f
的接口,所以我们可以轻松地将这些类与
solver\u t
类一起使用

下面是一个小程序来演示这一点

program test
  use solver
  use curves
  implicit none

  type(linear_curve), target :: linear
  type(polynomial_curve), target :: parabola
  type(solver_t) :: root_solver
  real :: root

  linear%m = 1.
  linear%b = 0.     ! y=x
  parabola%a = 1.
  parabola%b = 0.
  parabola%c = -1.  ! y=x^2-1

  root_solver%curve => linear
  root = root_solver%solve(-1., 1.)
  print *, "root  = ", root

  root_solver%curve => parabola
  root = root_solver%solve(-4., 0.5)
  print *, "root1 = ", root
  root = root_solver%solve(-0.5, 4.)
  print *, "root2 = ", root
end program test
这里我声明一些曲线,设置它们的参数,然后调用解算器来查找根。如果编译我们的曲线模块、测试程序并链接到我们先前创建的共享库,我们可以使用以下输出运行:

% ./roots                               
 root  =    0.00000000    
 root1 =   -1.00000286    
 root2 =    1.00000286

(根的质量受我放入第一个模块的示例解算器的质量限制,您可以做得更好)。这不是纯OO的最佳演示,因为solver_t类可以做得更好,但我重点演示了如何在编译solve_t时处理多条用户定义的曲线,而不必了解它们。

现代Fortran in Practice(Markus)一书中有这样的例子。我将阅读这本书,谢谢@Casey。
program test
  use solver
  use curves
  implicit none

  type(linear_curve), target :: linear
  type(polynomial_curve), target :: parabola
  type(solver_t) :: root_solver
  real :: root

  linear%m = 1.
  linear%b = 0.     ! y=x
  parabola%a = 1.
  parabola%b = 0.
  parabola%c = -1.  ! y=x^2-1

  root_solver%curve => linear
  root = root_solver%solve(-1., 1.)
  print *, "root  = ", root

  root_solver%curve => parabola
  root = root_solver%solve(-4., 0.5)
  print *, "root1 = ", root
  root = root_solver%solve(-0.5, 4.)
  print *, "root2 = ", root
end program test
% ./roots                               
 root  =    0.00000000    
 root1 =   -1.00000286    
 root2 =    1.00000286