Question

我试图更好地了解指针如何在fortran上工作以及如何可以用作其他函数的参数。该程序的工作方式如下：它具有三个条目r0,rf和long。然后，函数linspace生成一个从r0到rf的等距数组。然后，第二个函数应将半径值作为输入并取每个值的平方并生成一个新数组。这是代码

module class_Circle
 use, intrinsic :: iso_fortran_env, only: dp => real64
implicit none

 type, public :: Circle(long)
  integer,len :: long
  real(dp) :: r0,rf
 contains
  procedure  :: radius =>linspace
  procedure :: area => circle_area
 end type Circle

contains

 function linspace(this) result(radius)
  class(Circle(*)),intent(inout) :: this
  real(dp),dimension(this%long) :: radius
  integer :: i
  real(dp) :: incremento,b
  incremento = (this%rf-this%r0)/(1.0d0*(this%long-1))
  b = (this%r0*this%long-this%rf)/(1.0d0*(this%long-1))
  do i = 1,this%long
    radius(i) = b+incremento*i
  end do
 end function linspace

 function circle_area (this) result(area)
  class(Circle(*)), intent(inout) :: this
  real (dp), dimension(this%long):: area
  integer :: i
  real(dp), parameter :: pi =dacos(-1.0d0)
  do  i = 1,this%long 
    area(i) = pi * this%radius(i)**2
  end do
 end function circle_area

end module class_Circle

program circle_test
use, intrinsic :: iso_fortran_env, only: dp => real64
use class_Circle
implicit none

integer,parameter :: long = 3
type(Circle(long)) :: c
c%r0 = 1.5d0
c%rf = 3.0d0

print*, c%radius()
print*, c%area()
end program circle_test

编译gfortran-8时出现以下错误

      area(i) = pi * this%radius(i)**2.0d0
                               1
Error: ‘radius’ at (1) is not a member of the ‘Pdtcircle’ structure
objecttest03.f90:10:15:

   procedure :: area => circle_area
           1
Error: Argument ‘this’ of ‘circle_area’ with PASS(this) at (1) must be of        
the derived-type ‘circle' objecttest03.f90:9:15:

   procedure  :: radius =>linspace
           1
Error: Argument ‘this’ of ‘linspace’ with PASS(this) at (1) must be of the   derived-type ‘circle’

我不明白第一个错误，因为它是在类中定义的。另外，我想以下错误是由于这一原因造成的，那么我该如何进行这项工作？还有一个相关的问题，我可以指向不同模块中的过程吗？

如何在参数化派生类型中创建Fortran类型绑定过程？

0 个答案: