错误:具有非零等级的部分引用右侧的部分名称具有 ALLOCATABLE 属性 [英] Error: The part-name to the right of a part-ref with nonzero rank has the ALLOCATABLE attribute

问题描述

我想使用子例程 sum_real 访问数组派生类型中的数组元素.即:对所有人的权重中的第一个条目求和.

I would like to access the the elements of an array in a an arrayed derived type using the subroutine sum_real. That is: sum over first entry in the weight for all people.

type my_type
   real, dimension(:), allocatable :: weight
   real :: total_weight
end type my_type

type (my_type), dimension (:), allocatable :: people
type (my_type) :: answer

allocate (people (2))
allocate (people (1)%weight(2))
allocate (people (2)%weight(2))

people (1) % weight(1) = 1
people (2) % weight(1) = 1
people (1) % weight(2) = 3
people (2) % weight(2) = 3

call sum_real ( people (:) % weight(1), answer % total_weight )

我想要做的类似于在 派生类型数组中找到的答案:选择条目，除了我在数组派生类型中分配了一个数组而不是单个元素.

What I want to do is similar to the answer found in Array of derived type: select entry, except for the fact that I have an allocated array in an arrayed derived type instead of an single element.

但是，我得到一个编译器错误:

But, I get a compiler error:

error #7828: The part-name to the right of a part-ref with nonzero rank has the ALLOCATABLE attribute (6.1.2).   [WEIGHT]

推荐答案

如果您的组件是可分配的，那么您尝试的操作是不可能的.引用(6.1.2)其实是对官方标准文档的引用，禁止这样做.

What you try is not possible if your component is allocatable. The reference (6.1.2) is actually a reference to the official standard documents, which prohibits this.

原因很简单，可分配的组件(标量或数组)存储在与派生类型本身不同的内存部分.因此，如果你写

The reason is simple, the allocatable components (scalar or arrays) are stored in a different part of memory than the derived type itself. Therefore if you write

sum(people%total_weight)

或

people%total_weight = 0

没问题，total_weight 是不可分配的，它存储在派生类型中，编译器只是进入一个简单的循环并将一个接一个的字段设置为零.你可以预先知道每个%totalweight的地址.

it is no problem, total_weight is not allocatable, it is stored within the derived type and the compiler just goes in a simple loop and sets one field after another to zero. You can know the address of each %totalweight beforehand.

另一方面

sum(people%weight)

或

people%weight = 0

每个 %weight 都存储在其他地方，您没有任何简单的公式来计算每个 %weight(i) 的位置.

each %weight is stored elsewhere and you don't have any simple formula to compute where is each %weight(i).

如果可能的话，解决方案是固定数组的大小

The solution is either, to fix the size of the array, if possible

real, dimension(2) :: weight

或使用 do 循环

s = 0
do i = 1, size(people)
  S = S + sum(people(i)%weight)
end do

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