Calling C++ Modules in Python

[Stefan Behnel]

zxpatric at gmail.com, 11.03.2011 23:16:
> On Mar 11, 2011 4:59pm, Dan Stromberg wrote:
>
>> On Fri, Mar 11, 2011 at 1:15 PM, Patrick wrote:
>> I saw in the Beginner document that "•Is easily extended by adding new
>> modules implemented in a compiled language such as C or C++. ".
>> While to my investigation, it seems not that easy or did I miss
>> something?
>> boost python (C++ libraries need to be re-compiled with written
>> wrappers again?).
>>
>> SWIG (It works by taking the declarations found in C/C++ header and
>> using them to generate the wrapper code that scripting languages need
>> to access the underlying C/C++ code).
>>
>> I guess it should be asked: "easy for who"? "easy" is always relative to
>> some context.
>
>> If you're planning to stay with CPython and C++ forevermore, then Cython
>> is a nice way of gluing the two. Cython looks like Python code, but it
>> gives you access to C and C++ code and data as well as CPython code and
>> data.
>
> Thanks for the info. Really I was hoping for a "non-intrusive" way to
> expose existing C++ libraries to python. However, both solutions (BOOST,
> SWIG) listed here require the recompilation of libraries?! Seems Cython is
> doing the similar way?

Neither of the three requires the *re*compilation of libraries, but they 
all require you to compile wrapper code. As Dan pointed out, Cython has 
likely the lowest entry bar while giving you all freedom to design a 
suitable wrapper. It's also the fastest of the three.


> It is ok for small applications but will be hard for
> big software from big companies. The reason is that even companies are
> willing to rebuild their own libraries with those wrappers or decorations
> added(though needs approves from all sorts of departments :)), they can't
> force their 3rd-party developers/users.
> Correct me if I am wrong.

You can use the libraries as they are shipped, no need to change them.

Stefan