将Fortran代码集成到Julia中 [英] Integrating Fortran code in Julia

问题描述

我为自己的模块使用了GNU gfortran编译器（在Cygwin上）。一个很好的例子有望从编译阶段开始，解决出现错位的名称，并通过 ccall 从Julia调用子例程。我看过的大多数示例都跳过了前两个阶段。

所以想象一下，我在Fortran 90文件中命名为'f90tojl.f90'中有以下模块：

 模块m 
包含
整数函数five（）
 five = 5 
结束函数five 
 end module m 
  

这个例子来自 here 。我使用gfortran编译它以创建一个共享库：

  gfortran -shared -O2 f90tojl.f90 -o -fPIC f90tojl .so 
  

我的阅读茱莉亚文档的理解是，我应该能够像这样调用函数5：

pre $ ccall（（__ m_MOD_five，f90tojl），Int，（））

它并不适合我。我得到'错误编译匿名：无法加载模块f90tojl ... 。任何人都在意开导我吗？我得到了偷偷摸摸的感觉，我正在做一些愚蠢的事情。

在官方 doc ，重点在C.我也知道这是C ++的。在R和Python中，动力 - 我记得Cython和Rcpp--似乎是C / C ++。与此问题类似，我想了解一下如何将Julia与Fortran与Julia的接口与C / C ++。正如文档所说，调用Fortran和C基本上是一样的。我想，只有少得多的例子，因为人们想要包装的代码少得多。在Julia包装C是非常容易的，非常愉快。它通常不是为了加速Julia，就像使用R或Python一样，但更多的是利用已写好的代码。

Julia的标准库本身就是一个很好的例子与Fortran代码集成，例如这里是 Julia包装的.com / opencollab / ARPACK-NG> ARPARK 。在 Base 之外， glmnet 是用Fortran编写的，在那里是一个它的Julia包装（GLMNet.jl）。

我不明白现代Fortran区别如何。

I am using the GNU gfortran compiler (on Cygwin) for my own module. A good example will hopefully start from the compilation stage, address mangled names and call the subroutine from Julia via ccall. Most examples I've seen skip the first two stages.

So imagine that I have the following module in Fortran 90 file named 'f90tojl.f90':

module m
contains
    integer function five()
      five = 5
    end function five
end module m

This example is from here. I compile it with gfortran as follows to create a shared library:

gfortran -shared -O2 f90tojl.f90 -o -fPIC f90tojl.so

And my, admittedly shaky, understanding from reading the Julia docs suggest that I should be able to call the function five like so:

ccall( (:__m_MOD_five, "f90tojl"), Int, () )

It didn't work for me. I get 'error compiling anonymous: could not load module f90tojl.... Anyone cares to enlighten me? I got the sneaky sense I'm doing something silly....

In the official doc, the emphasis is on C. I'm also aware of this for C++. In R and Python, the momentum -- I have Cython and Rcpp in mind -- seems to be C/C++. Similar to this question, I want to get a sense of how easy it is to interface Julia with Fortran vs Julia with C/C++.

解决方案

Calling Fortran and C are essentially the same, as the documentation says. There are just far fewer examples because there is far less code people want to wrap, I suppose. Wrapping C is super easy in Julia, very pleasant. Its not typically done for speeding up Julia, like with R or Python, but more to take advantage of quality code already written.

Julia's standard library itself is a great example of integrating with Fortran code, e.g. here is the Julia wrapper for ARPARK. Outside of Base, glmnet is written in Fortran, and there is a Julia wrapper for it (GLMNet.jl).

I don't see how the modern Fortran distinction matters.

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