在使用type时如何将变量顺便设置成整型？

caimaxwell · 发表于 2023-3-8 17:11:18

大神们我又来了。代码如下：

[Fortran] 纯文本查看 复制代码

  type Subblocks
    integer*4  :: row
    integer*4  :: coL
  end type Subblocks

[Fortran] 纯文本查看 复制代码

  type blocks
    integer*4  :: row1,row2,coL1,coL2
    integer*4  :: rowm,coLm
    integer*4  :: dad
    integer*4  :: son(4)
    integer*4  :: bnum
    type(Subblocks), allocatable :: iSubblocks(:)
  end type blocks

[Fortran] 纯文本查看 复制代码

type(blocks), allocatable :: iblocks(:)

这段声明我是想完成这样一个操作：iblocks(i)%isubblocks(j)%coL

我想输出iblocks(i)、iblock(i)%isubblocks(j)、isubblocks(i)都是整数，想问问应该怎么声明呢？

caimaxwell · 发表于 2023-3-8 17:12:15

[Fortran] 纯文本查看 复制代码

type(Subblocks), allocatable  :: iSubblocks(:)

风平老涡 · 发表于 2023-3-8 21:26:06

本帖最后由风平老涡于 2023-3-8 23:38 编辑

caimaxwell 发表于 2023-3-8 17:12

[Fortran] 纯文本查看 复制代码

type Subblocks  
    integer*4  :: row 
    integer*4  :: coL
end type Subblocks

type blocks  
    integer*4  :: row1,row2,coL1,coL2 
    integer*4  :: rowm,coLm
    integer*4  :: dad
    integer*4  :: son(4)
    integer*4  :: bnum
    type(Subblocks), allocatable :: iSubblocks(:)
end type blocks

type(blocks), allocatable :: iblocks(:)
type(Subblocks), allocatable  :: iSubblocks(:)
integer :: i, n, m



n = 10
m = 20

allocate(iblocks(n))

do i = 1, n
   allocate(iblocks(i) % isubblocks(m))
end do

allocate(isubblocks(m))

print *, iblocks(3) % colm, iblocks % isubblocks(2) % col, isubblocks(3) % col

衍生类型的变量名（iblocks % isubblocks) 与变量名（isubblocks）不存在冲突。