如何初始化派生类型变量？

Question

如何初始化派生类型变量？

PS。我是一位老 F77 程序员，对使用 C# 的 OOP 有相当的了解，但对 Fortran 中的“派生类型”不熟悉。一名硕士生写了一大段代码就离开了。当使用 Intel 11（我不能再使用）编译时它可以工作，但使用 Intel 19（我可以在更大的项目中使用）时会抛出错误。我正在努力让它复活。 这是我在尝试理解学生代码时遇到的问题的 MWE。使用 Intel 11 不是一个选择。

program DerivedType
! investigate lines of code from large code "fabrics" that I do not understand

use strength_mod_test

! define and initialize array X
double precision, dimension(6) :: X = (/1,2,3,4,5,6/) ! just data, not PARAMETER. Values can be modified in the code.

! type (object) definition
type One  
    double precision :: small(6) ! the data in this type 
    ! pointer :: Ff(:,:) ! Pointer is not permitted as a statement within a derived-type-def.
end type One
type TxT
    double precision :: small(3,3) ! so that any var of type "type" has the correct size %small, depending of the "type" 
end type Txt
! Why the constructor is not inside the type definition. 
! Because "pointer" is not permitted inside a type def.
! You have to write the constructor outside the type def. 

! Variables must be defined before any executable code.  
! Define variables of type "type" using a constructor. 
! Cannot be defined inside the type if "pointer" is used. 
! "Type" constructor outside the type def. Note the () 
type(One), pointer :: Ff(:,:), Fw(:,:), sigmaf123(:,:), sigmaw123(:,:) 
type(TxT), pointer :: Df(:,:), Dw(:,:), Omega_f(:,:), Omega_w(:,:)

! local variables
integer i,j

! Initialize type variables
! Df is of type TxT, and in type TxT, small is (3,3)
! Df%small = 0 ! A component cannot be an array if the encompassing structure is an array.[SMALL]
! so I do this
do i=1,3
    do j=1,3
        Df(i,j)%small = 0.d0    ! why access violation?!
    enddo
end do
!    TYPE (TxT) :: Df_new = TxT (Df = 0)    does not work either

! I can't execute next statement until I initialize Df(i,j)%small and signf123(i,j)%small
Ff(i,j)%small = F_function (X, Df(i,j)%small, sigmaf123(i,j)%small)

end program

下面是模块strength_mod_test：

module strength_mod_test
contains
function F_function (R, D, sig)
    double precision, dimension(6) :: F_function            ! rank=1
    double precision, dimension(6), intent(in) :: R, sig    ! rank=1       
    double precision, dimension(3,3), intent(in) :: D       ! rank=2
    
    F_function = 0.d0
    
    if (D(1,1) == 0.d0 .AND. sig(1) > 0.d0) then
        F_function(1) = sig(1)/R(1)
    end if

    if (D(2,2) == 0.d0 .AND. sig(2) > 0.d0) then
        F_function(2) = sig(2)/R(2)
    end if
    
    if (D(3,3) == 0.d0 .AND. sig(3) > 0.d0) then
        F_function(3) = sig(3)/R(3)
    end if
    
    if (D(2,2) == 0.d0 .OR. D(3,3) == 0.d0) then
        F_function(4) = abs(sig(4))/R(4)
    end if
    
    if (D(3,3) == 0.d0) then
        F_function(5) = abs(sig(5))/R(5)
    end if
    
    if (D(2,2) == 0.d0) then
        F_function(6) = abs(sig(6))/R(6)
    end if
            
end function F_function
end module

Answer 1

感谢评论，很明显我忘记了分配指针，也忘记了正确初始化类型变量。下面的代码部分解决了所有这些问题。

    ! local variables
integer i,j
double precision, dimension(3,3) :: D_in=0.d0 ! used to initialize variables of type TxT, bec. the're 3x3 

! define and initialize array X
double precision, dimension(6) :: X = (/1,2,3,4,5,6/) ! just data, not PARAMETER. Values can be modified in the code.
integer sv ! to check if allocation/deallocation was successful

! Allocate and initialize type variables
allocate(Ff(nf, nw),STAT=sv);   if (sv /= 0) then; pause "sv/=0"; endif
allocate(Df(nf, nw),STAT=sv);   if (sv /= 0) then; pause "sv/=0"; endif
allocate(sigmaf123(nf, nw),STAT=sv);    if (sv /= 0) then; pause "sv/=0"; endif
! Df%small = 0 ! A component cannot be an array if the encompassing structure is an array.[SMALL]
! Solution: Df is of type TxT, & in type TxT, small is (3,3)
! Thus, each i,j element in Df must be initialized with a 3x3 matrix. 
do i=1,3        ! nf
    do j=1,3    ! nw
        Df(i,j)%small = D_in     ! 3x3 matrix of 3x3 matrices.
    enddo
end do

do i=1,3        ! nf
    do j=1,3    ! nw
        Ff(i,j)%small = F_function (X, Df(i,j)%small, sigmaf123(i,j)%small)
    enddo
enddo