How can I make a class look like a template with zero arguments?
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From: DeMarcus on 23 May 2010 03:46 Hi, To be able to initialize a member before any base class I have to inherit the member virtually as the first base class in the base-specifier-list. This is needed when dealing with std::streambuf. See http://accu.org/index.php/journals/264 Basically it looks like this. class MyOStream : private virtual MyStreamBuf, public std::ostream { public: MyOStream() : MyStreamBuf(), std::ostream( MyStreamBuf::buf() ) {} }; Now, by means of variadic templates in C++0x I have created a universal template that looks like this. template<template<typename...> class Object, typename... Tn> class InheritedComposition { public: template<typename... Args> InheritedComposition( Args... args ) : instance_( std::forward<Args>(args)... ) {} inline Object<Tn...>& instance() { return instance_; } private: Object<Tn...> instance_; }; I can do this. template<class T> class SomeTemplate { }; typedef InheritedComposition<SomeTemplate> ST; But I cannot do this. class SomeClass { }; typedef InheritedComposition<SomeClass> SC; How can I make SomeClass look like a template with zero arguments? Thanks, Daniel -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Paul Bibbings on 24 May 2010 00:45 DeMarcus <use_my_alias_here(a)hotmail.com> writes: <snip /> > But I cannot do this. > > class SomeClass > { > }; > > typedef InheritedComposition<SomeClass> SC; > > > How can I make SomeClass look like a template with zero arguments? Does the following achieve what you want? template< typename T = void > class SomeClass { }; typedef InheritedComposition< SomeClass > SC; I have to say that I'm not that up on variadic templates yet so I don't know what the implications of this idea are for your wider model. Regards Paul Bibbings -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Francis Glassborow on 24 May 2010 00:47 DeMarcus wrote: > I can do this. > > template<class T> > class SomeTemplate > { > }; > > typedef InheritedComposition<SomeTemplate> ST; > > > But I cannot do this. > > class SomeClass > { > }; > > typedef InheritedComposition<SomeClass> SC; > > > How can I make SomeClass look like a template with zero arguments? > > > Thanks, > Daniel > My instinct says that the answer lies in default template arguments. Now you want the default argument to be a type you would never use anywhere else so: struct empty_default_type {}; template<typename T = empty_defgault_type> SomeClass {}; Just an idea to play with. I don't have the time to fleash it out and think about the ramifications. -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: DeMarcus on 25 May 2010 01:34 On 2010-05-24 17:45, Paul Bibbings wrote: > DeMarcus<use_my_alias_here(a)hotmail.com> writes: > > <snip /> > >> But I cannot do this. >> >> class SomeClass >> { >> }; >> >> typedef InheritedComposition<SomeClass> SC; >> >> >> How can I make SomeClass look like a template with zero arguments? > > Does the following achieve what you want? > > template< typename T = void> > class SomeClass > { }; > > typedef InheritedComposition< SomeClass> SC; > > I have to say that I'm not that up on variadic templates yet so I don't > know what the implications of this idea are for your wider model. > > Regards > > Paul Bibbings > The thing is that I don't want to touch SomeClass at all since it could be an int. I would like to do some kind of "template cast". Like this. typedef InheritedComposition< template_cast< template<void> >( SomeClass ) > SC; Or if it could be done with a specialization of InheritedComposition; something like this (which doesn't work). template<class Object> class InheritedComposition { public: template<typename... Args> InheritedComposition( Args... args ) : instance_( std::forward<Args>(args)... ) {} inline Object& instance() { return instance_; } private: Object instance_; }; I'm just playing around trying to learn more about templates, especially about the variadic ones. My solution right now is the following where I changed the class name (but I'm not satisfied using a different name). template<class Object> class InheritedObjectComposition { public: template<typename... Args> InheritedComposition( Args... args ) : instance_( std::forward<Args>(args)... ) {} inline Object& instance() { return instance_; } private: Object instance_; }; I was wondering if there was a way to show the compiler that a class is a template with zero arguments. I think I found it! This would work, right? I can't try it since gcc 4.5 doesn't support template aliases yet. // Original universal template. template<template<typename...> class Object, typename... Tn> class InheritedComposition { public: template<typename... Args> InheritedComposition( Args... args ) : instance_( std::forward<Args>(args)... ) {} inline Object<Tn...>& instance() { return instance_; } private: Object<Tn...> instance_; }; // New template cast. template<typename T> using TemplateCast = T; // Typedef that didn't work before now works! typedef InheritedComposition< TemplateCast, SomeClass > SC; What do you think? Does your compiler support template aliases yet so you could try it out for me? -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Maxim Yegorushkin on 25 May 2010 01:36
On 23/05/10 19:46, DeMarcus wrote: > To be able to initialize a member before any base class I have to > inherit the member virtually as the first base class in the > base-specifier-list. Note that virtual is not required, you just need your member to be the first in the base class list. -- Max [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] |