一个关于继承的问题
module m_fatherimplicit none
type,abstract :: t_father
integer :: a
contains
procedure(equation),pass(this),deferred :: equation
end type t_father
abstract interface
module subroutine equation(this,other)
implicit none
class(t_father),intent(inout) :: this
class(t_father),intent(in) :: other
end subroutine
end interface
private equation
end module m_father
module m_son_1
use m_father
implicit none
type,extends(t_father) :: t_son_1
integer :: b
contains
procedure,pass(this) :: equation
procedure,pass(this) :: equation_own
generic,public :: assignment(=) => equation,equation_own
end type t_son_1
private equation
contains
subroutine equation(this,other)
implicit none
class(t_son_1),intent(inout) :: this
class(t_father),intent(in) :: other
this%a=other%a
end subroutine equation
subroutine equation_own(this,other)
implicit none
class(t_son_1),intent(inout) :: this
class(t_son_1),intent(in) :: other
this%a=other%a
this%b=other%b
end subroutine equation_own
end module m_son_1
module m_son_2
use m_father
implicit none
type,extends(t_father) :: t_son_2
integer :: c
contains
procedure,pass(this) :: equation
generic,public :: assignment(=) => equation
end type t_son_2
private equation
contains
subroutine equation(this,other)
implicit none
class(t_son_2),intent(inout) :: this
class(t_father),intent(in) :: other
this%a=other%a
end subroutine equation
end module m_son_2
program main
use m_son_1
use m_son_2
implicit none
type(t_son_1) :: a
type(t_son_2) :: b
b%a=1
a=b
call a%equation(b)
print *,a%a
end program main
我写的代码如上，目的是实现等号功能，
1、如果等号左边是某个继承类，而右边是同一个父类的其他继承类，则a想等。
2、如果等号左边是某个继承类，而右边是该继承类的子类，则a，b相等。
在加入红色字符之后，上面程序会引起冲突：Ambiguous interfaces in intrinsic assignment operator for ‘equation’ at (1) and ‘equation_own’ at (2)
还望组里大大不吝赐教，小弟十分感谢。
问题已经解决，使用select type... class is() ...class is() ...end select 即可 不吝赐教，希望学习一下改后的代码。
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