Current C++ compilers won't let you typedef a template, only a specialization. But, within the class, the template itself means "this particular specialization", so where it works it means the same thing. I'd use the more explicit form in the typedef, and the shortcut only if using it directly in a parameter list or something where it is used many times. --John
-----Original Message----- From: boost-users-bounces@lists.boost.org [mailto:boost-users- bounces@lists.boost.org] On Behalf Of Gao, Yi Sent: Wednesday, August 19, 2009 10:10 AM To: boost-users@lists.boost.org Subject: [Boost-users] shared_ptr on template class
Hi All,
//////In short: If I have a templated class A<T>, within it, I want to typedef shared_ptr to it self, should I write: typedef shared_ptr< A<T> > PointerType; or typedef shared_ptr< A > PointerType; ?
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