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I'm building a library for which one file requires an additional include path. Is there a way to adjust the include path for compilation of a single file?

     bld(features="cxx cxxshlib",
         source=[so, many, files, from an ant_glob],
         includes=[Some path that's really only needed for one interface file])

I'd be happy with a solution that is use based, too.

Ken White
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dbn
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2 Answers2

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I think most solutions will be more lines of code than just compiling your one file separately.

Inkrementator
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You need to compile the specfic file by using objects and then use the result.

Something like this:

def build(bld):
    # build the specfifc object
    bld.objects(source="foo.cpp", includess="path/to/directory", target="foo")
    # build the library and include that object file using 'use='
    bld.stlib(source='bla.cpp blu.cpp', includes="this/path that/path", target='mylibrary', use='foo')
user69453
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  • When I do this, the per-object use is skipped at the link step – dbn May 04 '23 at 16:49
  • If you don't link with the library, you need to use the ``use`` keyword and the object file to used as argument in your linking step also. – user69453 May 05 '23 at 18:55