On 4/21/06, Greg Ewing <[EMAIL PROTECTED]> wrote: ... > > GMP is covered by LGPL, so must any such derivative work > > But the wrapper is just using GMP as a library, so > it shouldn't be infected with LGPLness, should it?
If a lawyer for the PSF can confidently assert that gmpy is not a derivative work of GMP, I'll have no problem changing gmpy's licensing. But I won't make such a call myself: for example, gmpy.c #include's gmp.h and uses (==expands) some of the C macros there defined -- doesn't that make gmpy.o a derived work of gmp.h? I'm quite confident that the concept of "derived work" would not apply if gmpy.so only accessed a gmp.so (or other kinds of dynamic libraries), but I fear the connection is stronger than that, so, prudently, I'm assuming the "derived work" status until further notice. Alex _______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
