从其他数组继承大小的简洁表示法? [英] Concise notation for inheriting size from other array?

问题描述

在我的代码中，我有一个子例程，它接受一个 5 阶数组作为参数并使用一个局部变量，这是一个共享前 4 个索引的 4 阶数组.

In my code, I have a subroutine that takes a 5th-rank array as argument and uses a local variable, which is a 4-th rank array sharing the first 4 indices.

我正在尝试找到一种更简洁的方式来表达尺寸声明

I'm trying to find a more concise way to express the size declaration in

subroutine mysub(momentum)
  complex, intent(in) :: momentum(:,:,:,:,:)
  complex :: prefactor( &
      & size(momentum,1), size(momentum,2), size(momentum,4) &
      & size(momentum,5) )
  ...
end subroutine mysub

冗长的大小声明会损害可读性，尤其是当变量名比这里更长的时候.

The verbosity of the size declaration harms readability, especially when variable names are even longer than here.

如果这是 octave/matlab 我会预先分配 prefactor 通过编写

If this was octave/matlab I'd pre-allocate prefactor by writing

prefactor = zeros(size(momentum)([1 2 4 5]))

Fortran 90 是否支持类似简洁的东西?我知道可以使用预处理器宏来解决它，例如

Does Fortran 90 support something similarly concise? I know that it could be solved using preprocessor macros, such as

#define XSIZE2(array,a,b) SIZE(array,a), SIZE(array,b)
#define XSIZE3(array,a,b,c) SIZE(array,a), SIZE(array,b), SIZE(array,c)
#define XSIZE4(array,a,b,c,d) SIZE(array,a), SIZE(array,b), SIZE(array,c), SIZE(array,d)

但是引入这样的定义可能会损害可读性而不是帮助.

but introducing such definitions would probably harm the readability more than it helps.

推荐答案

Fortran 2008 在 allocate 语句中添加了 mold 说明符.如果您可以访问支持此功能的编译器，您可以尝试

Fortran 2008 added the mold specifier to the allocate statement. If you have access to a compiler that supports this feature, you can try

program main

  implicit none

  integer :: a(2,3,4,5,6)
  integer, allocatable :: b(:,:,:,:)

  print *, shape(a)

  allocate(b, mold=a(:,:,:,:,1))
  print *, shape(b)

end program main

此代码段适用于 Intel Fortran 2016，更新 1.

This snippet worked with Intel Fortran 2016, Update 1.

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