如何初始化Fortran派生类型变量？适配Intel 19编译问题

我熟悉F77、掌握C#面向对象编程，刚接触Fortran的派生类型（derived-type）。有一份研究生编写的代码在Intel 11编译器下可正常运行，但切换到Intel 19编译器后报错，问题出在初始化派生类型指针变量时出现访问违规，无法执行后续代码。以下是最小可复现示例（MWE）：

主程序代码 #

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

模块代码 #

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

核心问题是指针变量未关联到有效的内存空间：Intel 11编译器对未初始化指针的检查较宽松，而Intel 19编译器严格遵循Fortran标准，直接访问未关联的指针会触发访问违规。

具体修复步骤 #

1. 修正派生类型名称拼写错误：主程序中定义type TxT，但结束时写成end type Txt（大小写错误），这会导致编译错误，需修正为end type TxT。

2. 为指针分配有效内存：
   Fortran的指针必须先关联到已分配的内存或目标变量才能使用，不能直接像普通变量一样赋值。推荐两种修复方式：

方式一：用可分配数组替代指针（简单高效，无需指针特性时优先选择） #

program DerivedType
use strength_mod_test

! define and initialize array X
double precision, dimension(6) :: X = (/1,2,3,4,5,6/)

! type definition
type One  
    double precision :: small(6)
end type One
type TxT
    double precision :: small(3,3)
end type TxT  ! 修正拼写错误

! 使用可分配数组替代指针，无需手动管理关联
type(One), allocatable :: Ff(:,:), sigmaf123(:,:) 
type(TxT), allocatable :: Df(:,:)

! local variables
integer i,j

! 为可分配数组分配内存（维度根据实际需求调整）
allocate(Ff(3,3), sigmaf123(3,3))
allocate(Df(3,3))

! Initialize type variables
do i=1,3
    do j=1,3
        Df(i,j)%small = 0.d0
        sigmaf123(i,j)%small = 0.d0  ! 初始化sigmaf123，避免后续访问未初始化数据
    enddo
end do

! 执行函数调用（循环内操作，避免原代码的数组越界问题）
do i=1,3
    do j=1,3
        Ff(i,j)%small = F_function(X, Df(i,j)%small, sigmaf123(i,j)%small)
    enddo
end do

! 释放内存（程序结束时编译器会自动释放，显式释放更规范）
deallocate(Ff, sigmaf123, Df)

end program

方式二：保留指针并关联到目标内存 #

program DerivedType
use strength_mod_test

double precision, dimension(6) :: X = (/1,2,3,4,5,6/)

type One  
    double precision :: small(6)
end type One
type TxT
    double precision :: small(3,3)
end type TxT

! 定义目标数组（带target属性）
type(One), target :: Ff_target(3,3), sigmaf123_target(3,3)
type(TxT), target :: Df_target(3,3)
! 定义指针并关联到目标
type(One), pointer :: Ff(:,:), sigmaf123(:,:) 
type(TxT), pointer :: Df(:,:)

integer i,j

! 关联指针到目标内存
Ff => Ff_target
sigmaf123 => sigmaf123_target
Df => Df_target

! 初始化
do i=1,3
    do j=1,3
        Df(i,j)%small = 0.d0
        sigmaf123(i,j)%small = 0.d0
    enddo
end do

! 函数调用
do i=1,3
    do j=1,3
        Ff(i,j)%small = F_function(X, Df(i,j)%small, sigmaf123(i,j)%small)
    enddo
end do

end program

补充说明 #

• Fortran 90及以后标准支持派生类型内部包含指针，原代码中“Pointer is not permitted as a statement within a derived-type-def”的注释错误，只是含指针的派生类型无法用默认构造函数初始化指针成员，需手动关联或自定义构造函数。
• 原代码循环结束后直接使用i=4,j=4的值访问数组，会触发越界，修复后的代码改为嵌套循环内调用函数，避免该问题。

内容的提问来源于stack exchange，提问作者Echeban