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[optional] Why is operator-> disabled
by Jiri Palecek 26 Jun '06

26 Jun '06

Hello, I have been using boost::optional with a reference parameter recently, and I was puzzled by the fact that it is not possible to use operator-> to get the content. I could understand that get_pointer is disabled, but there is always a pointer such as *p==optional.get() and you can't actually get the pointer from operator-> Regards Jiri Palecek

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Re: [Boost-users] Nindi ... Boost Variant
by Ovanes Markarian 26 Jun '06

26 Jun '06

The operator << must be defined in the boost namespace. See the error message of VC:see reference to function template instantiation 'std::basic_ostream<_Elem,_Traits> &boost::operator <<<char,std::char_traits<char>,std::vector<_Ty>,int,std::string,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_,boost::detail::variant::void_>(std::basic_ostream<_Elem,_Traits> &,const boost::variant<T0_,T1,T2> &)' being compiled VC looks for boost::operator<< function and not ::operator<< function. But this should be documented in OutputStreamable. This would compile: #include<iostream> #include<vector> #include<string> #include<boost/variant.hpp> namespace boost { template <typename T> std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) { for(typename std::vector<T>::size_type i = 0; i < v.size(); ++i) os << v[i] << std::endl; return os; } } struct TpTemplateVariant { typedef boost::variant<std::vector<unsigned long>,int, std::string> BoostVariant; }; typedef TpTemplateVariant::BoostVariant Tpvariant; int main () { Tpvariant p(5); std::cout << p; return 0; } Good Luck, Ovanes Markarian On Mon, June 26, 2006 12:45, CLAIRE, Narinder, Group Risk Mgmt wrote: > I apologise for emailing you directly .. but i canno access the mailing list > from work.. > However in response to your response. > > > The following doesn't compile either .. I get the error > > error C2679: binary '<<' : no operator found which takes a right-hand > operand of type 'const T0' (or there is no acceptable conversion) > > On VC8 > > #include<iostream> > #include<vector> > #include<string> > > #include<boost/variant.hpp> > > > > std::basic_ostream<char> &operator<<(std::basic_ostream<char> > &theostream,const std::vector<unsigned long> &theVector) { > for(unsigned long i(0);i<theVector.size();i++) > theostream << theVector[i] << "\n" ; > return theostream; > } > > > struct TpTemplateVariant { > typedef boost::variant< std::vector<unsigned > long>, > > int, > > std::string > >> BoostVariant; > }; > > typedef TpTemplateVariant::BoostVariant Tpvariant; > > int main () { > > Tpvariant p(5); > std::cout << p; > > return 0; > > > > > *********************************************************************************** > The Royal Bank of Scotland plc. Registered in Scotland No 90312. Registered Office: 36 St Andrew > Square, Edinburgh EH2 2YB. > Authorized and regulated by the Financial Services Authority > > This e-mail message is confidential and for use by the > addressee only. If the message is received by anyone other > than the addressee, please return the message to the sender > by replying to it and then delete the message from your > computer. Internet e-mails are not necessarily secure. The > Royal Bank of Scotland plc does not accept responsibility for > changes made to this message after it was sent. > > Whilst all reasonable care has been taken to avoid the > transmission of viruses, it is the responsibility of the recipient to > ensure that the onward transmission, opening or use of this > message and any attachments will not adversely affect its > systems or data. No responsibility is accepted by The Royal > Bank of Scotland plc in this regard and the recipient should carry > out such virus and other checks as it considers appropriate. > Visit our websites at: > http://www.rbos.com > http://www.rbsmarkets.com > *********************************************************************************** >

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Issue with program_options with STLPort using long descriptions
by Tim 26 Jun '06

26 Jun '06

Dear All, I have an issue when using program_options with long description strings when using STLport in debug mode. This occurs within format_paragraph in options_description.cpp and is being found by the _Incrementable check within the STLPort iterator checking. This is using vc71 STLPort 5.0 and Boost 1.33.1 Any help on this issue would be gratefully received. Apologies for asking a question that has been asked previously but there was no resolution posted to the group.

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Strange compiler behaviour with mpl::vector
by Ovanes Markarian 26 Jun '06

26 Jun '06

Hello all! I posted the following message http://lists.boost.org/boost-users/2006/06/20271.php a few days ago, but did not get any replies. I know there is an error in the const_ref class. But nevertheless the problem which occures is that my class accepts a vector of supported types as template parameter. It has two constructors (no default ctor): template<class Seq_> class my_class { public: typedef my_class<Seq_> this_type; explicit inline my_class(const this_type& copy_from) :... {...} template<class TypeFromVector> explicit inline my_class(typename const_ref<TypeFromVector>::type inst) :... {...} }; Now if I try to initialize my_class with type declared in vector I get an error: error C2664: 'discriminated_union<Seq_>::discriminated_union(const discriminated_union<Seq_> &)' : cannot convert parameter 1 from 't1' to 'const discriminated_union<Seq_> &' with [ Seq_=types_vector ] Reason: cannot convert from 't1' to 'const discriminated_union<Seq_>' with [ Seq_=types_vector ] No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called Ok, if I define the ctor param without typename const_ref<...>::type everything goes fine, if I use const_ref it fails. const_ref is defined as follows: template<class Ty> struct const_ref : boost::add_reference < typename boost::add_const < typename boost::remove_reference<Ty>::type >::type > {}; I can understand that the compiler is unable to make a decision which ctor to use, but I do not get an ambiguaty error. It seems like the compiler simply quits expanding templates and uses the last most successful ctor matched before the error. This happens only when I use mpl::vector templated type in conjunction with templated ctor. Other types like below work fine: template<int D_, class Ty_> struct my_pair : mpl::pair<mpl::int_<D_>, Ty_> { typedef my_pair<D_, Ty_> my_type; inline my_pair() {} explicit inline my_pair( typename const_ref<Ty_>::type ty) : second_(ty) {} explicit inline my_pair(const my_type& other) : second_(other.second_) {} inline my_type& operator =(const my_type& other) { second_=other.second_; return *this; } second second_; }; May be some of you have any suggestions. Thanks a lot, Ovanes Markarian

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Serialization of an object with multiple formats
by Vidal Roca 26 Jun '06

26 Jun '06

It is possible to define more than one serialization format for a data structure with the serialization library? If not, has someone some good purposes to make it to? Thanks, Vidal

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Re: [Boost-users] Boost Variant OutputStreamable
by Ovanes Markarian 26 Jun '06

26 Jun '06

Hi! Can you please to compile this code either with string in the variant or a vector. Not sure if it helps but the variant documentation at http://www.boost.org/doc/html/variant/reference.html states: namespace boost { template<typename T1, typename T2 = unspecified, ..., typename TN = unspecified> class variant; template<typename Sequence> class make_variant_over; template<typename T1, typename T2, ..., typename TN> void swap(variant<T1, T2, ..., TN> &, variant<T1, T2, ..., TN> &); template<typename ElemType, typename Traits, typename T1, typename T2, ..., typename TN> std::basic_ostream<ElemType,Traits> & operator<<(std::basic_ostream<ElemType,Traits> &, const variant<T1, T2, ..., TN> &); } Which means that the types are not zero-index based. So your T5 would be a string and not a vector. Make some tests like with simple integers, strings and finally a vector and see what happens. Best Regards, Ovanes -----Original Message----- From: Nindi [mailto:Nindi73@yahoo.co.uk] Sent: Sunday, June 25, 2006 15:36 To: boost-users(a)lists.boost.org Subject: [Boost-users] Boost Variant OutputStreamable I have started using Boost recently, and though I should switch my own variant to the Boost Variant . However I cannot compile the following code #include<iostream> #include<vector> #include<string> #include<boost/variant.hpp> std::basic_ostream<char> &operator<<(std::basic_ostream<char> &theostream,const std::vector<unsigned long> &theVector) { for(unsigned long i(0);i<theVector.size();i++) theostream << theVector[i] << "\n" ; return theostream; } struct TpTemplateVariant { typedef boost::variant< int, unsigned long, double, bool, std::string, std::vector<unsigned long> > BoostVariant; }; typedef TpTemplateVariant::BoostVariant Tpvariant; int main () { Tpvariant p(5); std::cout << p; return 0; } The error I get is error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'const T5'

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[BGL] total newbie question
by pixelpusher＠cox.net 25 Jun '06

25 Jun '06

Sorry for the newbieishness of the question, but I'm just getting started with C++ templates in general and Boost.Graph in particular. It seems to me that, while there is plenty of it, the documentation assumes a level of understanding of templates and other OO principles that I simply don't have at this point. Normally, I'd start digging into the example programs and look for common points of reference with things I already know and use the comments in the code for guidance. However, the example programs for BGL are astonishingly devoid of comments. They seem to be written for folks who already know how BGL works, not for folks who are trying to learn it. Most of what I see in the examples leaves me scratching my head and wondering WTF am I getting myself into here. Anyway, what I was hoping to find was more of a tutorial in how to get started writing a path-finding program using BGL and (probably) the Dijkstra-shortest paths template. - I have a set of edges that I will read from a text file. I don't know in advance how many edges will be read, so I plan to use an std::vector to store each edge's id, nodes, and weight. (All of the examples construct an array of known size and use that to build the graph. None use variable sized structures to build the graph.) Unless, of course, I can load the graph without pre-constructing a storage scheme. - The edge weights have to be modified based on the user's preferences, which will change with every run. (None of the examples show how to change edge weights.) - I will have known starting and ending nodes. I don't need to know the paths from my source to every possible destination. I only need to extract the best one that leads to my destination. I'd be happy enough for now just to know how to get started on the first bullet. Can anyone provide any guidance here? I'd really appreciate it. Thanks, Thom -- ------------------------------------------------------- Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue. -- Winston Churchill

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Re: [Boost-users] Strange compiler behaviour with mpl::vector
by Ovanes Markarian 25 Jun '06

25 Jun '06

Sorry, in the error message you will read the type name discriminated_union. This is the real type name, think of it as my_class. Thanks and sorry again, Ovanes _____ From: Ovanes Markarian [mailto:om_boost@keywallet.com] Sent: Sunday, June 25, 2006 23:18 To: 'boost-users(a)lists.boost.org' Subject: Strange compiler behaviour with mpl::vector Hello all! I posted the following message http://lists.boost.org/boost-users/2006/06/20271.php a few days ago, but did not get any replies. I know there is an error in the const_ref class. But nevertheless the problem which occures is that my class accepts a vector of supported types as template parameter. It has two constructors (no default ctor): template<class Seq_> class my_class { public: typedef my_class<Seq_> this_type; explicit inline my_class(const this_type& copy_from) :... {...} template<class TypeFromVector> explicit inline my_class(typename const_ref<TypeFromVector>::type inst) :... {...} }; Now if I try to initialize my_class with type declared in vector I get an error: error C2664: 'discriminated_union<Seq_>::discriminated_union(const discriminated_union<Seq_> &)' : cannot convert parameter 1 from 't1' to 'const discriminated_union<Seq_> &' with [ Seq_=types_vector ] Reason: cannot convert from 't1' to 'const discriminated_union<Seq_>' with [ Seq_=types_vector ] No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called Ok, if I define the ctor param without typename const_ref<...>::type everything goes fine, if I use const_ref it fails. const_ref is defined as follows: template<class Ty> struct const_ref : boost::add_reference < typename boost::add_const < typename boost::remove_reference<Ty>::type >::type > {}; I can understand that the compiler is unable to make a decision which ctor to use, but I do not get an ambiguaty error. It seems like the compiler simply quits expanding templates and uses the last most successful ctor matched before the error. This happens only when I use mpl::vector templated type in conjunction with templated ctor. Other types like below work fine: template<int D_, class Ty_> struct my_pair : mpl::pair<mpl::int_<D_>, Ty_> { typedef my_pair<D_, Ty_> my_type; inline my_pair() {} explicit inline my_pair( typename const_ref<Ty_>::type ty) : second_(ty) {} explicit inline my_pair(const my_type& other) : second_(other.second_) {} inline my_type& operator =(const my_type& other) { second_=other.second_; return *this; } second second_; }; May be some of you have any suggestions. Thanks a lot, Ovanes Markarian

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Boost Variant OutputStreamable
by Nindi 25 Jun '06

25 Jun '06

I have started using Boost recently, and though I should switch my own variant to the Boost Variant . However I cannot compile the following code #include<iostream> #include<vector> #include<string> #include<boost/variant.hpp> std::basic_ostream<char> &operator<<(std::basic_ostream<char> &theostream,const std::vector<unsigned long> &theVector) { for(unsigned long i(0);i<theVector.size();i++) theostream << theVector[i] << "\n" ; return theostream; } struct TpTemplateVariant { typedef boost::variant< int, unsigned long, double, bool, std::string, std::vector<unsigned long> > BoostVariant; }; typedef TpTemplateVariant::BoostVariant Tpvariant; int main () { Tpvariant p(5); std::cout << p; return 0; } The error I get is error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'const T5'

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Re: [Boost-users] inspecting templates and boost::is_pointer
by Michael Nicolella 24 Jun '06

24 Jun '06

You passed has_xxx to the macro - It would make sense to only pass xxx (looks like an error in the example): BOOST_MPL_HAS_XXX_TRAIT_DEF(xxx) Also, you'll need to instantiate the has_xxx template: typedef has_xxx<test2> my_test; Mike -----Original Message----- From: boost-users-bounces(a)lists.boost.org [mailto:boost-users-bounces@lists.boost.org] On Behalf Of Ovanes Markarian Sent: Saturday, June 24, 2006 12:22 PM To: boost-users(a)lists.boost.org Subject: Re: [Boost-users] inspecting templates and boost::is_pointer Dave, thanks for your answer. I think this should be definitely a part of FAQ, otherwise someone could spent endless hours on matters why some of boost classes do not work as expected. I have one more question. I tried the following code with VC 8.0: #include <boost/mpl/has_xxx.hpp> struct test2 { typedef int xxx; }; BOOST_MPL_HAS_XXX_TRAIT_DEF(has_xxx) typedef has_xxx my_test; And got a compiler error: main.cpp(22) : error C2146: syntax error : missing ';' before identifier 'my_test' Somehow the macro was not expanded. I replaced xxx with some other name before but end up in compiler errors, so that I decided to check if the standard example is compilable at all by using the typedef. Do you have any suggestions? Thanks, Ovanes -----Original Message----- From: David Abrahams [mailto:dave@boost-consulting.com] Sent: Saturday, June 24, 2006 04:51 To: boost-users(a)lists.boost.org Subject: Re: [Boost-users] inspecting templates and boost::is_pointer "Ovanes Markarian" <om_boost(a)keywallet.com> writes: > Hello all, > > I am trying to specialize boost::is_pointer so, that it can determine > if a type is a pointer wrapper (shared_ptr, auto_ptr, weak_ptr). First of all, don't do that. It's a really bad idea, and it will probably break some (Boost) library code. is_pointer is supposed to detect real pointers, and if you start changing what it does with other types, you'll break the expectations of every bit of code that currently uses is_pointer. This has come up before, so maybe it should be a FAQ (John?) > Does anybody has ideas how this can be done? May be it would be > possible to handle if e.g. one can determine if a class/struct has > some typedef inside: http://www.boost.org/libs/mpl/doc/refmanual/has-xxx-trait-def.html -- Dave Abrahams Boost Consulting www.boost-consulting.com _______________________________________________ Boost-users mailing list Boost-users(a)lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users

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