Question

我遇到Fortran析构函数/最终例程的问题。有下面的代码：

  module my_type_module
    implicit none

    type my_type
      contains
        final :: destructor
    end type my_type

    interface my_type
      ! Constructor declaration
      module procedure my_type_constructor
    end interface my_type

    contains
      function my_type_constructor()
        type(my_type) :: my_type_constructor

        print *, 'In constructor address is: ',
 &          loc(my_type_constructor)

      end function my_type_constructor

      subroutine destructor(this)
        type(my_type) :: this
        print *, 'Destructor of my_type object with address: ',
 &        loc(this)
      end subroutine destructor

  end module my_type_module

  program trial

    use my_type_module
    implicit none

    type(my_type) :: argument

    print *, 'Trial program starts'
    print *, 'Initial address is', loc(argument)

    argument = my_type_constructor()

    print *, 'Address afer constructor is called is', loc(argument)

    print *, 'doing some work...'
    print *, 'finishing program...'

    print *, 'Final address is', loc(argument)

  end program trial

输出是：

Trial program starts
Initial address is            4351590240
In constructor address is:        140734743834256
Destructor of my_type object with address: 4351590240
Destructor of my_type object with address: 140734743834256
Address afer constructor is called is      4351590240     
doing some work...
finishing program...
Final address is            4351590240

因此，似乎构造对象在构造结束时而不是在程序结束时被破坏。有什么想法有什么不对吗？上面的代码是用ifort 14.0.0编译的，

Answer 1

没有错（明显（？）在2013年使用固定格式源继续使用Fortran 2003代码。）

第一个“析构函数”调用是完成赋值语句的左侧。

第二个调用是最终确定函数结果（概念上，在表达式中使用了该结果的值并将其传送到赋值的左侧）。

在程序终止之前存在的实体（此处执行结束程序声明）尚未最终确定。

Fortran最终例程在变量超出范围之前调用自身

1 个答案: