手柄数组派生类型 [英] Handle on array derived type

问题描述

我想访问数组的元素使用子程序sum_real一个一个阵列派生类型。这就是：在为所有的人在举重第一项之和。

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 )

我想要做的是类似于派生类型的阵列找到了答案：选择进入，除非我有在阵列派生类型，而不是一个单一元素的数组分配的事实。

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.

不过，我得到一个编译器错误：

But, I get a compiler error:

错误＃7828：该部分名称的一部分，裁判有非零秩的右边有ALLOCATABLE属性（6.1.2）。 【重量】

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

在尝试什么是不可能的，如果你的组件是可分配的。参考（ 6.1.2 ）实际上是官方标准文件的引用，禁止这一点。

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，分配组件（标量或阵列）存储在内存比派生类型本身的不同部分。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

是 total_weight 没有问题，是不是分配的，它存储派生类型中，编译器只是去在一个简单的循环，并设置一个字段陆续零。你可以知道每一个地址％totalweight 事先

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.

在另一方面

sum(people%weight)

或

people%weight = 0

每个％重量存储在别处，你没有任何简单的公式来计算，其中为每个％重量（I）。

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

的溶液可以是，固定阵列的大小，如果可能的话

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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