On Monday, 10 de October de 2011 13:34:21 Simon Hausmann wrote: > > It has to because by construction B's classes are used in A, so you have > > to > > use B to use A. > > Not quite, it doesn't mean that you can do #include <ClassFromB> if you do > QT += A.
But shouldn't it? If I have:
#include <ClassFromA>
which in turn has:
#include <B/ClassFromB>
Then I am using that class from B. More importantly, if I use that class from
B, then my application has to link to B *directly*, so the buildsystem must
know this dependency.
--
Thiago Macieira - thiago (AT) macieira.info - thiago (AT) kde.org
Software Architect - Intel Open Source Technology Center
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