Fortran 数组在添加值时自动增长 [英] Fortran array automatically growing when adding a value

问题描述

是否有任何现有方法可以在 Fortran 中模拟不断增长的数组?就像 C++ 中的向量.当我在 Internet 上没有找到有关此主题的任何内容时，我感到非常惊讶.

Is there any existing way to emulate growing array in Fortran? Like vector in C++. I was very surprised when I haven't found anything on this subject on the Internet.

作为一个动机示例，假设我计算了一些递推关系并且我想存储我得到的所有中间数.我的停止标准是相邻结果之间的差异，所以我无法事先知道应该为此分配多少内存.

As a motivation example, suppose I compute some recurrence relation and I want to store all the intermediate numbers I get. My stopping criterion is the difference between adjacent results so I cannot know beforehand how much memory I should allocate for this.

推荐答案

我确信它以前曾在本网站的某个地方展示过，但我找不到.

I am sure it has been shown somewhere on this site before, but I cannot find it.

首先，在 Fortran 2003 中，您可以通过简单的方式添加一个元素

First, in Fortran 2003, you can add one element by simple

a = [a, item]

正如弗朗西斯卡卢斯评论的那样.这很可能会经常重新分配数组，而且速度会很慢.

as commented by francescalus. This is likely to reallocate the array very often and will be slow.

您可以将数组分配到比元素数n 更大的大小.当您的元素数量 n 增长到超过数组 size(a) 的大小时，您可以分配一个更大的新数组(此处为 2x)并复制旧元素那里.不幸的是，Fortran 中没有 realloc().

You can keep your array to be allocated to somewhat larger size then your number of elements n. When your number of elements n grows above the size of the array size(a) you can allocate a new array larger by some factor (here 2x) and copy the old elements there. There is no realloc() in Fortran, unfortunately.

module growing_array
  implicit none

  real, allocatable :: a(:)

  integer :: n

contains

  subroutine add_item(item)
    real, allocatable :: tmp(:)
    real, intent(in) :: item

    if (n == size(a)) then
      !this statement is F2003, it can be avoided, but I don't see why in 2016
      call move_alloc(a, tmp)

      allocate(a(n*2))
      a(1:n) = tmp
    end if

    n = n + 1

    a(n) = item
  end subroutine
end module

我省略了初始分配，很简单.

I left out the initial allocation, it is simple enough.

这一切都可以放在带有类型绑定过程的派生类型中，并将其用作数据结构，但那是纯 Fortran 2003 而你想要 90.所以我展示 Fortran 95，因为 Fortran 90 在许多方面存在缺陷可分配数组的方法，已经过时了，基本上已经死了.

It all can be put into a derived type with type-bound procedures, and use it as a data structure, but that is pure Fortran 2003 and you wanted 90. So I show Fortran 95, because Fortran 90 is flawed in many ways for allocatable arrays and is desperately obsolete and essentially dead.

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