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; ? //////In more detail I have: template< typename T> class A { public: typedef shared_ptr< A<T> > PointerType; // or typedef shared_ptr< A > PointerType; // Further more static PointerType New() { return PointerType(new A<T>); // or return PointerType(new A); } }; Could I know if there's difference in between? Thanks! Yi Gao
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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Probably a bit offtopic, but I do the following when working with templated classes: typedef MyClass<P1> me; typedef boost::shared_ptr<me> pointer; so, later everything will depend on "me", for example: return pointer(new me); Also, this code could be reused when creating other classes.
typedef MyClass<P1> me; typedef boost::shared_ptr<me> pointer;
so, later everything will depend on "me", for example: return pointer(new me); Also, this code could be reused when creating other classes.
Note that this is not necessary: template <typename T> struct Foo { Foo foobar (); }; template <typename T> Foo<T> Foo<T>::foobar () { return Foo(); // << referred to Foo<T> implicitly } int main () { Foo<void> foo; Foo<void> phoo = foo.foobar(); } Inside the class declaration and inside corpora of member functions, the unqualified class name is an alias for the qualified one. Or did I misunderstood something? Sebastian Mach http://phresnel.org
participants (4)
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Gao, Yi -
John Dlugosz -
Roman Shmelev -
Sebastian Mach