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  • Function signature-like expressions as C++ template arguments

    - by Jeff Lee
    I was looking at Don Clugston's FastDelegate mini-library and noticed a weird syntactical trick with the following structure: TemplateClass< void( int, int ) > Object; It almost appears as if a function signature is being used as an argument to a template instance declaration. This technique (whose presence in FastDelegate is apparently due to one Jody Hagins) was used to simplify the declaration of template instances with a semi-arbitrary number of template parameters. To wit, it allowed this something like the following: // A template with one parameter template<typename _T1> struct Object1 { _T1 m_member1; }; // A template with two parameters template<typename _T1, typename _T2> struct Object2 { _T1 m_member1; _T2 m_member2; }; // A forward declaration template<typename _Signature> struct Object; // Some derived types using "function signature"-style template parameters template<typename _Dummy, typename _T1> struct Object<_Dummy(_T1)> : public Object1<_T1> {}; template<typename _Dummy, typename _T1, typename _T2> struct Object<_Dummy(_T1, _T2)> : public Object2<_T1, _T2> {}; // A. "Vanilla" object declarations Object1<int> IntObjectA; Object2<int, char> IntCharObjectA; // B. Nifty, but equivalent, object declarations typedef void UnusedType; Object< UnusedType(int) > IntObjectB; Object< UnusedType(int, char) > IntCharObjectB; // C. Even niftier, and still equivalent, object declarations #define DeclareObject( ... ) Object< UnusedType( __VA_ARGS__ ) > DeclareObject( int ) IntObjectC; DeclareObject( int, char ) IntCharObjectC; Despite the real whiff of hackiness, I find this kind of spoofy emulation of variadic template arguments to be pretty mind-blowing. The real meat of this trick seems to be the fact that I can pass textual constructs like "Type1(Type2, Type3)" as arguments to templates. So here are my questions: How exactly does the compiler interpret this construct? Is it a function signature? Or, is it just a text pattern with parentheses in it? If the former, then does this imply that any arbitrary function signature is a valid type as far as the template processor is concerned? A follow-up question would be that since the above code sample is valid code, why doesn't the C++ standard just allow you to do something like the following, which is does not compile? template<typename _T1> struct Object { _T1 m_member1; }; // Note the class identifier is also "Object" template<typename _T1, typename _T2> struct Object { _T1 m_member1; _T2 m_member2; }; Object<int> IntObject; Object<int, char> IntCharObject;

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  • GCC error with variadic templates: "Sorry, unimplemented: cannot expand 'Identifier...' into a fixe

    - by Dennis
    While doing variadic template programming in C++0x on GCC, once in a while I get an error that says "Sorry, unimplemented: cannot expand 'Identifier...' into a fixed-length arugment list." If I remove the "..." in the code then I get a different error: "error: parameter packs not expanded with '...'". So if I have the "..." in, GCC calls that an error, and if I take the "..." out, GCC calls that an error too. The only way I have been able to deal with this is to completely rewrite the template metaprogram from scratch using a different approach, and (with luck) I eventually come up with code that doesn't cause the error. But I would really like to know what I was doing wrong. Despite Googling for it and despite much experimentation, I can't pin down what it is that I'm doing differently between variadic template code that does produce this error, and code that does not have the error. The wording of the error message seems to imply that the code should work according the C++0x standard, but that GCC doesn't support it yet. Or perhaps it is a compiler bug? Here's some code that produces the error. Note: I don't need you to write a correct implementation for me, but rather just to point out what is about my code that is causing this specific error // Used as a container for a set of types. template <typename... Types> struct TypePack { // Given a TypePack<T1, T2, T3> and T=T4, returns TypePack<T1, T2, T3, T4> template <typename T> struct Add { typedef TypePack<Types..., T> type; }; }; // Takes the set (First, Others...) and, while N > 0, adds (First) to TPack. // TPack is a TypePack containing between 0 and N-1 types. template <int N, typename TPack, typename First, typename... Others> struct TypePackFirstN { // sorry, unimplemented: cannot expand ‘Others ...’ into a fixed-length argument list typedef typename TypePackFirstN<N-1, typename TPack::template Add<First>::type, Others...>::type type; }; // The stop condition for TypePackFirstN: when N is 0, return the TypePack that has been built up. template <typename TPack, typename... Others> struct TypePackFirstN<0, TPack, Others...> //sorry, unimplemented: cannot expand ‘Others ...’ into a fixed-length argument list { typedef TPack type; }; EDIT: I've noticed that while a partial template instantiation that looks like does incur the error: template <typename... T> struct SomeStruct<1, 2, 3, T...> {}; Rewriting it as this does not produce an error: template <typename... T> struct SomeStruct<1, 2, 3, TypePack<T...>> {}; It seems that you can declare parameters to partial specializations to be variadic; i.e. this line is OK: template <typename... T> But you cannot actually use those parameter packs in the specialization, i.e. this part is not OK: SomeStruct<1, 2, 3, T... The fact that you can make it work if you wrap the pack in some other type, i.e. like this: SomeStruct<1, 2, 3, TypePack<T...>> to me implies that the declaration of the variadic parameter to a partial template specialization was successful, and you just can't use it directly. Can anyone confirm this?

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  • Extracting pair member in lambda expressions and template typedef idiom

    - by Nazgob
    Hi, I have some complex types here so I decided to use nifty trick to have typedef on templated types. Then I have a class some_container that has a container as a member. Container is a vector of pairs composed of element and vector. I want to write std::find_if algorithm with lambda expression to find element that have certain value. To get the value I have to call first on pair and then get value from element. Below my std::find_if there is normal loop that does the trick. My lambda fails to compile. How to access value inside element which is inside pair? I use g++ 4.4+ and VS 2010 and I want to stick to boost lambda for now. #include <vector> #include <algorithm> #include <boost\lambda\lambda.hpp> #include <boost\lambda\bind.hpp> template<typename T> class element { public: T value; }; template<typename T> class element_vector_pair // idiom to have templated typedef { public: typedef std::pair<element<T>, std::vector<T> > type; }; template<typename T> class vector_containter // idiom to have templated typedef { public: typedef std::vector<typename element_vector_pair<T>::type > type; }; template<typename T> bool operator==(const typename element_vector_pair<T>::type & lhs, const typename element_vector_pair<T>::type & rhs) { return lhs.first.value == rhs.first.value; } template<typename T> class some_container { public: element<T> get_element(const T& value) const { std::find_if(container.begin(), container.end(), bind(&typename vector_containter<T>::type::value_type::first::value, boost::lambda::_1) == value); /*for(size_t i = 0; i < container.size(); ++i) { if(container.at(i).first.value == value) { return container.at(i); } }*/ return element<T>(); //whatever } protected: typename vector_containter<T>::type container; }; int main() { some_container<int> s; s.get_element(5); return 0; }

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  • C++0x: How can I access variadic tuple members by index at runtime?

    - by nonoitall
    I have written the following basic Tuple template: template <typename... T> class Tuple; template <uintptr_t N, typename... T> struct TupleIndexer; template <typename Head, typename... Tail> class Tuple<Head, Tail...> : public Tuple<Tail...> { private: Head element; public: template <uintptr_t N> typename TupleIndexer<N, Head, Tail...>::Type& Get() { return TupleIndexer<N, Head, Tail...>::Get(*this); } uintptr_t GetCount() const { return sizeof...(Tail) + 1; } private: friend struct TupleIndexer<0, Head, Tail...>; }; template <> class Tuple<> { public: uintptr_t GetCount() const { return 0; } }; template <typename Head, typename... Tail> struct TupleIndexer<0, Head, Tail...> { typedef Head& Type; static Type Get(Tuple<Head, Tail...>& tuple) { return tuple.element; } }; template <uintptr_t N, typename Head, typename... Tail> struct TupleIndexer<N, Head, Tail...> { typedef typename TupleIndexer<N - 1, Tail...>::Type Type; static Type Get(Tuple<Head, Tail...>& tuple) { return TupleIndexer<N - 1, Tail...>::Get(*(Tuple<Tail...>*) &tuple); } }; It works just fine, and I can access elements in array-like fashion by using tuple.Get<Index() - but I can only do that if I know the index at compile-time. However, I need to access elements in the tuple by index at runtime, and I won't know at compile-time which index needs to be accessed. Example: int chosenIndex = getUserInput(); cout << "The option you chose was: " << tuple.Get(chosenIndex) << endl; What's the best way to do this?

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  • calling template function without <>; type inference

    - by Oops
    Hi, if I have a function template with typename T, where the compiler can set the type by itself, I do not have to write the type explicitely when I call the function like: template < typename T > T min( T v1, T v2 ) { return ( v1 < v2 ) ? v1: v2; } int i1 = 1, i2 = 2; int i3 = min( i1, i2 ); //no explicit <type> but if I have a function template with two different typenames like... template < typename TOut, typename TIn > TOut round( TIn v ) { return (TOut)( v + 0.5 ); } double d = 1.54; int i = round<int>(d); //explicit <int> Is it true that I have to specify at least 1 typename, always? I assume the reason is because C++ can not distinguish functions between different return types, true? but if I use a void function and handover a reference, again I must not explicitely specify the return typename: template < typename TOut, typename TIn > void round( TOut & vret, TIn vin ) { vret = (TOut)(vin + 0.5); } double d = 1.54; int i; round(i, d); //no explicit <int> should the conclusion be to avoid functions with return and more prefer void functions that return via a reference when writing templates? Or is there a possibility to avoid explicitely writing the return type? something like "type inference" for templates... is "type inference" possible in C++0x? I hope I was not too unclear. many thanks in advance Oops

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  • C++ Template Classes Circular Dependency Problem

    - by TomWij
    We have two classes: template<typename T, typename Size, typename Stack, typename Sparse> class Matrix and template<typename T, typename Size> class Iterator Matrix should be able to return begin and end iterators and Iterator will keep a referrence to the Matrix to access the elements via it's interface. We don't want Iterator to depend on the internal storage of the Matrix to prevent coupling. How can we solve this cyclic dependency problem? (The internal Storage class has the same template parameters as the Matrix class and the same access procedures as the Matrix itself)

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  • What are possible suffixes after variable name in VBA?

    - by archimed7592
    As I've already figured out, there is at least six of them: !@#$%&. Here is snip: Dim A!, B@, C#, D$, E%, F& Debug.Print "A! - " & TypeName(A) Debug.Print "B@ - " & TypeName(B) Debug.Print "C# - " & TypeName(C) Debug.Print "D$ - " & TypeName(D) Debug.Print "E% - " & TypeName(E) Debug.Print "F& - " & TypeName(F) Outputs A! - Single B@ - Currency C# - Double D$ - String E% - Integer F& - Long Where is documentation on this syntax sugar? What other possible suffixes are there? Is there one for Date?

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  • Fluent NHibernate auto mapping: map property from another table's column

    - by queen3
    I'm trying to use S#arp architecture... which includes Fluent NHibernate I'm newbie with (and with NHibernate too, frankly speaking). Auto mapping is used. So I have this: public class UserAction : Entity { public UserAction() { } [DomainSignature] [NotNull, NotEmpty] public virtual string Name { get; set; } [NotNull, NotEmpty] public virtual string TypeName { get; private set; } } public class UserActionMap : IAutoMappingOverride<UserAction> { public void Override(AutoMap<UserAction> mapping) { mapping.WithTable("ProgramComponents", m => m.Map(x => x.TypeName)); } } Now, table UserActions references table ProgramComponents (many to one) and I want property UserAction.TypeName to have value from db field ProgramComponents.TypeName. However, the above code fails with NHibernate.MappingException: Duplicate property mapping of TypeName found in OrderEntry3.Core.UserAction As far as I understand the problem is that TypeName is already auto-mapped... but I haven't found a way to remove the automatic mapping. Actually I think that my WithTable/Map mapping has to replace the automatic TypeName mapping, but seems like it does not. I also tried different mapping (names are different but that's all the same): mapping.WithTable("ProgramComponents", m => m.References<ProgramComponent>( x => x.Selector, "ProductSelectorID" ) and still get the same error. I can overcome this with mapping.HasOne<ProgramComponent>(x => x.Selector); but that's not what I exactly wants to do. And I still wonder why the first two methods do not work. I suspect this is because of WithTable.

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  • Should this work?

    - by Noah Roberts
    I am trying to specialize a metafunction upon a type that has a function pointer as one of its parameters. The code compiles just fine but it will simply not match the type. #include <iostream> #include <boost/mpl/bool.hpp> #include <boost/mpl/identity.hpp> template < typename CONT, typename NAME, typename TYPE, TYPE (CONT::*getter)() const, void (CONT::*setter)(TYPE const&) > struct metafield_fun {}; struct test_field {}; struct test { int testing() const { return 5; } void testing(int const&) {} }; template < typename T > struct field_writable : boost::mpl::identity<T> {}; template < typename CONT, typename NAME, typename TYPE, TYPE (CONT::*getter)() const > struct field_writable< metafield_fun<CONT,NAME,TYPE,getter,0> > : boost::mpl::false_ {}; typedef metafield_fun<test, test_field, int, &test::testing, 0> unwritable; int main() { std::cout << typeid(field_writable<unwritable>::type).name() << std::endl; std::cin.get(); } Output is always the type passed in, never bool_.

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  • Difference between std::result_of and decltype

    - by Luc Touraille
    I have some trouble understanding the need for std::result_of in C++0x. If I understood correctly, result_of is used to obtain the resulting type of invoking a function object with certain types of parameters. For example: template <typename F, typename Arg> typename std::result_of<F(Arg)> invoke(F f, Arg a) { return f(a); } I don't really see the difference with the following code: template <typename F, typename Arg> auto invoke(F f, Arg a) -> decltype(f(a)) //uses the f parameter { return f(a); } or template <typename F, typename Arg> auto invoke(F f, Arg a) -> decltype(F()(a)); //"constructs" an F { return f(a); } The only problem I can see with these two solutions is that we need to either: have an instance of the functor to use it in the expression passed to decltype. know a defined constructor for the functor. Am I right in thinking that the only difference between decltype and result_of is that the first one needs an expression whereas the second does not?

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  • GCC ICE -- alternative function syntax, variadic templates and tuples

    - by Marc H.
    (Related to C++0x, How do I expand a tuple into variadic template function arguments?.) The following code (see below) is taken from this discussion. The objective is to apply a function to a tuple. I simplified the template parameters and modified the code to allow for a return value of generic type. While the original code compiles fine, when I try to compile the modified code with GCC 4.4.3, g++ -std=c++0x main.cc -o main GCC reports an internal compiler error (ICE) with the following message: main.cc: In function ‘int main()’: main.cc:53: internal compiler error: in tsubst_copy, at cp/pt.c:10077 Please submit a full bug report, with preprocessed source if appropriate. See <file:///usr/share/doc/gcc-4.4/README.Bugs> for instructions. Question: Is the code correct? or is the ICE triggered by illegal code? // file: main.cc #include <tuple> // Recursive case template<unsigned int N> struct Apply_aux { template<typename F, typename T, typename... X> static auto apply(F f, const T& t, X... x) -> decltype(Apply_aux<N-1>::apply(f, t, std::get<N-1>(t), x...)) { return Apply_aux<N-1>::apply(f, t, std::get<N-1>(t), x...); } }; // Terminal case template<> struct Apply_aux<0> { template<typename F, typename T, typename... X> static auto apply(F f, const T&, X... x) -> decltype(f(x...)) { return f(x...); } }; // Actual apply function template<typename F, typename T> auto apply(F f, const T& t) -> decltype(Apply_aux<std::tuple_size<T>::value>::apply(f, t)) { return Apply_aux<std::tuple_size<T>::value>::apply(f, t); } // Testing #include <string> #include <iostream> int f(int p1, double p2, std::string p3) { std::cout << "int=" << p1 << ", double=" << p2 << ", string=" << p3 << std::endl; return 1; } int g(int p1, std::string p2) { std::cout << "int=" << p1 << ", string=" << p2 << std::endl; return 2; } int main() { std::tuple<int, double, char const*> tup(1, 2.0, "xxx"); std::cout << apply(f, tup) << std::endl; std::cout << apply(g, std::make_tuple(4, "yyy")) << std::endl; } Remark: If I hardcode the return type in the recursive case (see code), then everything is fine. That is, substituting this snippet for the recursive case does not trigger the ICE: // Recursive case (hardcoded return type) template<unsigned int N> struct Apply_aux { template<typename F, typename T, typename... X> static int apply(F f, const T& t, X... x) { return Apply_aux<N-1>::apply(f, t, std::get<N-1>(t), x...); } }; Alas, this is an incomplete solution to the original problem.

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  • Parameter pack argument consumption

    - by yuri kilochek
    It is possible to get the first element of the parameter pack like this template <typename... Elements> struct type_list { }; template <typename TypeList> struct type_list_first_element { }; template <typename FirstElement, typename... OtherElements> struct type_list_first_element<type_list<FirstElement, OtherElements...>> { typedef FirstElement type; }; int main() { typedef type_list<int, float, char> list; typedef type_list_first_element<list>::type element; return 0; } but not possible to similary get the last element like this template <typename... Elements> struct type_list { }; template <typename TypeList> struct type_list_last_element { }; template <typename LastElement, typename... OtherElements> struct type_list_last_element<type_list<OtherElements..., LastElement>> { typedef LastElement type; }; int main() { typedef type_list<int, float, char> list; typedef type_list_last_element<list>::type element; return 0; } with gcc 4.7.1 complaining: error: 'type' in 'struct type_list_last_element<type_list<int, float, char>>' does not name a type What paragraps from the standard describe this behaviour? It seems to me that template parameter packs are greedy in a sense that they consume all matching arguments, which in this case means that OtherElements consumes all three arguments (int, float and char) and then there is nothing left for LastElement so the compilation fails. Am i correct in the assumption? EDIT: To clarify: I am not asking how to extract the last element from the parameter pack, i know how to do that. What i actually want is to pick the pack apart from the back as opposed to the front, and as such recursing all the way to the back for each element would be ineffective. Apparentely reversing the sequence beforehand is the most sensible choice.

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  • Templated derived class in CRTP (Curiously Recurring Template Pattern)

    - by Butterwaffle
    Hi, I have a use of the CRTP that doesn't compile with g++ 4.2.1, perhaps because the derived class is itself a template? Does anyone know why this doesn't work or, better yet, how to make it work? Sample code and the compiler error are below. Source: foo.C #include <iostream> using namespace std; template<typename X, typename D> struct foo; template<typename X> struct bar : foo<X,bar<X> > { X evaluate() { return static_cast<X>( 5.3 ); } }; template<typename X> struct baz : foo<X,baz<X> > { X evaluate() { return static_cast<X>( "elk" ); } }; template<typename X, typename D> struct foo : D { X operator() () { return static_cast<D*>(this)->evaluate(); } }; template<typename X, typename D> void print_foo( foo<X,D> xyzzx ) { cout << "Foo is " << xyzzx() << "\n"; } int main() { bar<double> br; baz<const char*> bz; print_foo( br ); print_foo( bz ); return 0; } Compiler errors foo.C: In instantiation of ‘foo<double, bar<double> >’: foo.C:8: instantiated from ‘bar<double>’ foo.C:30: instantiated from here foo.C:18: error: invalid use of incomplete type ‘struct bar<double>’ foo.C:8: error: declaration of ‘struct bar<double>’ foo.C: In instantiation of ‘foo<const char*, baz<const char*> >’: foo.C:13: instantiated from ‘baz<const char*>’ foo.C:31: instantiated from here foo.C:18: error: invalid use of incomplete type ‘struct baz<const char*>’ foo.C:13: error: declaration of ‘struct baz<const char*>’

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  • Feedback on iterating over type-safe enums

    - by Sumant
    In response to the earlier SO question "Enumerate over an enum in C++", I came up with the following reusable solution that uses type-safe enum idiom. I'm just curious to see the community feedback on my solution. This solution makes use of a static array, which is populated using type-safe enum objects before first use. Iteration over enums is then simply reduced to iteration over the array. I'm aware of the fact that this solution won't work if the enumerators are not strictly increasing. template<typename def, typename inner = typename def::type> class safe_enum : public def { typedef typename def::type type; inner val; static safe_enum array[def::end - def::begin]; static bool init; static void initialize() { if(!init) // use double checked locking in case of multi-threading. { unsigned int size = def::end - def::begin; for(unsigned int i = 0, j = def::begin; i < size; ++i, ++j) array[i] = static_cast<typename def::type>(j); init = true; } } public: safe_enum(type v = def::begin) : val(v) {} inner underlying() const { return val; } static safe_enum * begin() { initialize(); return array; } static safe_enum * end() { initialize(); return array + (def::end - def::begin); } bool operator == (const safe_enum & s) const { return this->val == s.val; } bool operator != (const safe_enum & s) const { return this->val != s.val; } bool operator < (const safe_enum & s) const { return this->val < s.val; } bool operator <= (const safe_enum & s) const { return this->val <= s.val; } bool operator > (const safe_enum & s) const { return this->val > s.val; } bool operator >= (const safe_enum & s) const { return this->val >= s.val; } }; template <typename def, typename inner> safe_enum<def, inner> safe_enum<def, inner>::array[def::end - def::begin]; template <typename def, typename inner> bool safe_enum<def, inner>::init = false; struct color_def { enum type { begin, red = begin, green, blue, end }; }; typedef safe_enum<color_def> color; template <class Enum> void f(Enum e) { std::cout << static_cast<unsigned>(e.underlying()) << std::endl; } int main() { std::for_each(color::begin(), color::end(), &f<color>); color c = color::red; }

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  • Using Unity – Part 6

    - by nmarun
    This is the last of the ‘Unity’ series and I’ll be talking about generics here. If you’ve been following the previous articles, you must have noticed that I’m just adding more and more ‘Product’ classes to the project. I’ll change that trend in this blog where I’ll be adding an ICaller interface and a Caller class. 1: public interface ICaller<T> where T : IProduct 2: { 3: string CallMethod<T>(string typeName); 4: } 5:  6: public class Caller<T> : ICaller<T> where T:IProduct 7: { 8: public string CallMethod<T>(string typeName) 9: { 10: //... 11: } 12: } We’ll fill-in the implementation of the CallMethod in a few, but first, here’s what we’re going to do: create an instance of the Caller class pass it the IProduct as a generic parameter in the CallMethod method, we’ll use Unity to dynamically create an instance of IProduct implemented object I need to add the config information for ICaller and Caller types. 1: <typeAlias alias="ICaller`1" type="ProductModel.ICaller`1, ProductModel" /> 2: <typeAlias alias="Caller`1" type="ProductModel.Caller`1, ProductModel" /> The .NET Framework’s convention to express generic types is ICaller`1, where the digit following the "`" matches the number of types contained in the generic type. So a generic type that contains 4 types contained in the generic type would be declared as: 1: <typeAlias alias="Caller`4" type="ProductModel.Caller`4, ProductModel" /> On my .aspx page, I have the following UI design: 1: <asp:RadioButton ID="LegacyProduct" Text="Product" runat="server" GroupName="ProductWeb" 2: AutoPostBack="true" OnCheckedChanged="RadioButton_CheckedChanged" /> 3: <br /> 4: <asp:RadioButton ID="NewProduct" Text="Product 2" runat="server" GroupName="ProductWeb" 5: AutoPostBack="true" OnCheckedChanged="RadioButton_CheckedChanged" /> 6: <br /> 7: <asp:RadioButton ID="ComplexProduct" Text="Product 3" runat="server" GroupName="ProductWeb" 8: AutoPostBack="true" OnCheckedChanged="RadioButton_CheckedChanged" /> 9: <br /> 10: <asp:RadioButton ID="ArrayConstructor" Text="Product 4" runat="server" GroupName="ProductWeb" 11: AutoPostBack="true" OnCheckedChanged="RadioButton_CheckedChanged" /> Things to note here are that all these radio buttons belong to the same GroupName => only one of these four can be clicked. Next, all four controls postback to the same ‘OnCheckedChanged’ event and lastly the ID’s point to named types of IProduct (already added to the web.config file). 1: <type type="IProduct" mapTo="Product" name="LegacyProduct" /> 2:  3: <type type="IProduct" mapTo="Product2" name="NewProduct" /> 4:  5: <type type="IProduct" mapTo="Product3" name="ComplexProduct"> 6: ... 7: </type> 8:  9: <type type="IProduct" mapTo="Product4" name="ArrayConstructor"> 10: ... 11: </type> In my calling code, I see which radio button was clicked, pass that as an argument to the CallMethod method. 1: protected void RadioButton_CheckedChanged(object sender, EventArgs e) 2: { 3: string typeName = ((RadioButton)sender).ID; 4: ICaller<IProduct> caller = unityContainer.Resolve<ICaller<IProduct>>(); 5: productDetailsLabel.Text = caller.CallMethod<IProduct>(typeName); 6: } What’s basically happening here is that the ID of the control gets passed on to the typeName which will be one of “LegacyProduct”, “NewProduct”, “ComplexProduct” or “ArrayConstructor”. I then create an instance of an ICaller and pass the typeName to it. Now, we’ll fill in the blank for the CallMethod method (sorry for the naming guys). 1: public string CallMethod<T>(string typeName) 2: { 3: IUnityContainer unityContainer = HttpContext.Current.Application["UnityContainer"] as IUnityContainer; 4: T productInstance = unityContainer.Resolve<T>(typeName); 5: return ((IProduct)productInstance).WriteProductDetails(); 6: } This is where I’ll resolve the IProduct by passing the type name and calling the WriteProductDetails() method. With all things in place, when I run the application and choose different radio buttons, the output should look something like below:          Basically this is how generics come to play in Unity. Please see the code I’ve used for this here. This marks the end of the ‘Unity’ series. I’ll definitely post any updates that I find, but for now I don’t have anything planned.

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  • Specializing a template member function of a template class?

    - by uj2
    I have a template class that has a template member function that needs to be specialized, as in: template <typename T> class X { public: template <typename U> void Y() {} template <> void Y<int>() {} }; Altough VC handles this correctly, apperantly this isn't standard and GCC complains: explicit specialization in non-namespace scope 'class X<T>' I tried: template <typename T> class X { public: template <typename U> void Y() {} }; template <typename T> // Also tried `template<>` here void X<T>::Y<int>() {} But this causes both VC and GCC to complain. What's the right way to do this?

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  • Odd behavior when recursively building a return type for variadic functions

    - by Dennis Zickefoose
    This is probably going to be a really simple explanation, but I'm going to give as much backstory as possible in case I'm wrong. Advanced apologies for being so verbose. I'm using gcc4.5, and I realize the c++0x support is still somewhat experimental, but I'm going to act on the assumption that there's a non-bug related reason for the behavior I'm seeing. I'm experimenting with variadic function templates. The end goal was to build a cons-list out of std::pair. It wasn't meant to be a custom type, just a string of pair objects. The function that constructs the list would have to be in some way recursive, with the ultimate return value being dependent on the result of the recursive calls. As an added twist, successive parameters are added together before being inserted into the list. So if I pass [1, 2, 3, 4, 5, 6] the end result should be {1+2, {3+4, 5+6}}. My initial attempt was fairly naive. A function, Build, with two overloads. One took two identical parameters and simply returned their sum. The other took two parameters and a parameter pack. The return value was a pair consisting of the sum of the two set parameters, and the recursive call. In retrospect, this was obviously a flawed strategy, because the function isn't declared when I try to figure out its return type, so it has no choice but to resolve to the non-recursive version. That I understand. Where I got confused was the second iteration. I decided to make those functions static members of a template class. The function calls themselves are not parameterized, but instead the entire class is. My assumption was that when the recursive function attempts to generate its return type, it would instantiate a whole new version of the structure with its own static function, and everything would work itself out. The result was: "error: no matching function for call to BuildStruct<double, double, char, char>::Go(const char&, const char&)" The offending code: static auto Go(const Type& t0, const Type& t1, const Types&... rest) -> std::pair<Type, decltype(BuildStruct<Types...>::Go(rest...))> My confusion comes from the fact that the parameters to BuildStruct should always be the same types as the arguments sent to BuildStruct::Go, but in the error code Go is missing the initial two double parameters. What am I missing here? If my initial assumption about how the static functions would be chosen was incorrect, why is it trying to call the wrong function rather than just not finding a function at all? It seems to just be mixing types willy-nilly, and I just can't come up with an explanation as to why. If I add additional parameters to the initial call, it always burrows down to that last step before failing, so presumably the recursion itself is at least partially working. This is in direct contrast to the initial attempt, which always failed to find a function call right away. Ultimately, I've gotten past the problem, with a fairly elegant solution that hardly resembles either of the first two attempts. So I know how to do what I want to do. I'm looking for an explanation for the failure I saw. Full code to follow since I'm sure my verbal description was insufficient. First some boilerplate, if you feel compelled to execute the code and see it for yourself. Then the initial attempt, which failed reasonably, then the second attempt, which did not. #include <iostream> using std::cout; using std::endl; #include <utility> template<typename T1, typename T2> std::ostream& operator <<(std::ostream& str, const std::pair<T1, T2>& p) { return str << "[" << p.first << ", " << p.second << "]"; } //Insert code here int main() { Execute(5, 6, 4.3, 2.2, 'c', 'd'); Execute(5, 6, 4.3, 2.2); Execute(5, 6); return 0; } Non-struct solution: template<typename Type> Type BuildFunction(const Type& t0, const Type& t1) { return t0 + t1; } template<typename Type, typename... Rest> auto BuildFunction(const Type& t0, const Type& t1, const Rest&... rest) -> std::pair<Type, decltype(BuildFunction(rest...))> { return std::pair<Type, decltype(BuildFunction(rest...))> (t0 + t1, BuildFunction(rest...)); } template<typename... Types> void Execute(const Types&... t) { cout << BuildFunction(t...) << endl; } Resulting errors: test.cpp: In function 'void Execute(const Types& ...) [with Types = {int, int, double, double, char, char}]': test.cpp:33:35: instantiated from here test.cpp:28:3: error: no matching function for call to 'BuildFunction(const int&, const int&, const double&, const double&, const char&, const char&)' Struct solution: template<typename... Types> struct BuildStruct; template<typename Type> struct BuildStruct<Type, Type> { static Type Go(const Type& t0, const Type& t1) { return t0 + t1; } }; template<typename Type, typename... Types> struct BuildStruct<Type, Type, Types...> { static auto Go(const Type& t0, const Type& t1, const Types&... rest) -> std::pair<Type, decltype(BuildStruct<Types...>::Go(rest...))> { return std::pair<Type, decltype(BuildStruct<Types...>::Go(rest...))> (t0 + t1, BuildStruct<Types...>::Go(rest...)); } }; template<typename... Types> void Execute(const Types&... t) { cout << BuildStruct<Types...>::Go(t...) << endl; } Resulting errors: test.cpp: In instantiation of 'BuildStruct<int, int, double, double, char, char>': test.cpp:33:3: instantiated from 'void Execute(const Types& ...) [with Types = {int, int, double, double, char, char}]' test.cpp:38:41: instantiated from here test.cpp:24:15: error: no matching function for call to 'BuildStruct<double, double, char, char>::Go(const char&, const char&)' test.cpp:24:15: note: candidate is: static std::pair<Type, decltype (BuildStruct<Types ...>::Go(BuildStruct<Type, Type, Types ...>::Go::rest ...))> BuildStruct<Type, Type, Types ...>::Go(const Type&, const Type&, const Types& ...) [with Type = double, Types = {char, char}, decltype (BuildStruct<Types ...>::Go(BuildStruct<Type, Type, Types ...>::Go::rest ...)) = char] test.cpp: In function 'void Execute(const Types& ...) [with Types = {int, int, double, double, char, char}]': test.cpp:38:41: instantiated from here test.cpp:33:3: error: 'Go' is not a member of 'BuildStruct<int, int, double, double, char, char>'

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  • Serializing object with no namespaces using DataContractSerializer

    - by Yurik
    How do I remove XML namespaces from an object's XML representation serialized using DataContractSerializer? That object needs to be serialized to a very simple output XML. Latest & greatest - using .Net 4 beta 2 The object will never need to be deserialized. XML should not have any xmlns:... namespace refs Any subtypes of Exception and ISubObject need to be supported. It will be very difficult to change the original object. Object: [Serializable] class MyObj { string str; Exception ex; ISubObject subobj; } Need to serialize into: <xml> <str>...</str> <ex i:nil="true" /> <subobj i:type="Abc"> <AbcProp1>...</AbcProp1> <AbcProp2>...</AbcProp2> </subobj> </xml> I used this code: private static string ObjectToXmlString(object obj) { if (obj == null) throw new ArgumentNullException("obj"); var serializer = new DataContractSerializer( obj.GetType(), null, Int32.MaxValue, false, false, null, new AllowAllContractResolver()); var sb = new StringBuilder(); using (var xw = XmlWriter.Create(sb, new XmlWriterSettings { OmitXmlDeclaration = true, NamespaceHandling = NamespaceHandling.OmitDuplicates, Indent = true })) { serializer.WriteObject(xw, obj); xw.Flush(); return sb.ToString(); } } From this article I adopted a DataContractResolver so that no subtypes have to be declared: public class AllowAllContractResolver : DataContractResolver { public override bool TryResolveType(Type dataContractType, Type declaredType, DataContractResolver knownTypeResolver, out XmlDictionaryString typeName, out XmlDictionaryString typeNamespace) { if (!knownTypeResolver.TryResolveType(dataContractType, declaredType, null, out typeName, out typeNamespace)) { var dictionary = new XmlDictionary(); typeName = dictionary.Add(dataContractType.FullName); typeNamespace = dictionary.Add(dataContractType.Assembly.FullName); } return true; } public override Type ResolveName(string typeName, string typeNamespace, Type declaredType, DataContractResolver knownTypeResolver) { return knownTypeResolver.ResolveName(typeName, typeNamespace, declaredType, null) ?? Type.GetType(typeName + ", " + typeNamespace); } }

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  • Binder and variadic template ends up in a segmentation fault

    - by phlipsy
    I wrote the following program #include <iostream> template<typename C, typename Res, typename... Args> class bind_class_t { private: Res (C::*f)(Args...); C *c; public: bind_class_t(Res (C::*f)(Args...), C* c) : f(f), c(c) { } Res operator() (Args... args) { return (c->*f)(args...); } }; template<typename C, typename Res, typename... Args> bind_class_t<C, Res, Args...> bind_class(Res (C::*f)(Args...), C* c) { return bind_class<C, Res, Args...>(f, c); } class test { public: int add(int x, int y) { return x + y; } }; int main() { test t; // bind_class_t<test, int, int, int> b(&test::add, &t); bind_class_t<test, int, int, int> b = bind_class(&test::add, &t); std::cout << b(1, 2) << std::endl; return 0; } compiled it with gcc 4.3.3 and got a segmentation fault. After spending some time with gdb and this program it seems to me that the addresses of the function and the class are mixed up and a call of the data address of the class isn't allowed. Moreover if I use the commented line instead everything works fine. Can anyone else reproduce this behavior and/or explain me what's going wrong here?

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  • Overlapping template partial specialization when wanting an "override" case: how to avoid the error?

    - by user173342
    I'm dealing with a pretty simple template struct that has an enum value set by whether its 2 template parameters are the same type or not. template<typename T, typename U> struct is_same { enum { value = 0 }; }; template<typename T> struct is_same<T, T> { enum { value = 1 }; }; This is part of a library (Eigen), so I can't alter this design without breaking it. When value == 0, a static assert aborts compilation. So I have a special numerical templated class SpecialCase that can do ops with different specializations of itself. So I set up an override like this: template<typename T> struct SpecialCase { ... }; template<typename LT, typename RT> struct is_same<SpecialCase<LT>, SpecialCase<RT>> { enum { value = 1 }; }; However, this throws the error: more than one partial specialization matches the template argument list Now, I understand why. It's the case where LT == RT, which steps on the toes of is_same<T, T>. What I don't know is how to keep my SpecialCase override and get rid of the error. Is there a trick to get around this? edit: To clarify, I need all cases where LT != RT to also be considered the same (have value 1). Not just LT == RT.

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  • Operator+ for a subtype of a template class.

    - by baol
    I have a template class that defines a subtype. I'm trying to define the binary operator+ as a template function, but the compiler cannot resolve the template version of the operator+. #include <iostream> template<typename other_type> struct c { c(other_type v) : cs(v) {} struct subtype { subtype(other_type v) : val(v) {} other_type val; } cs; }; template<typename other_type> typename c<other_type>::subtype operator+(const typename c<other_type>::subtype& left, const typename c<other_type>::subtype& right) { return typename c<other_type>::subtype(left.val + right.val); } // This one works // c<int>::subtype operator+(const c<int>::subtype& left, // const c<int>::subtype& right) // { return c<int>::subtype(left.val + right.val); } int main() { c<int> c1 = 1; c<int> c2 = 2; c<int>::subtype cs3 = c1.cs + c2.cs; std::cerr << cs3.val << std::endl; } I think the reason is because the compiler (g++4.3) cannot guess the template type so it's searching for operator+<int> instead of operator+. What's the reason for that? What elegant solution can you suggest?

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  • Operator+ for a subtype of a template classe.

    - by baol
    I have a template class that defines a subtype. I'm trying to define the binary operator+ as a template function, but the compiler cannot resolve the template version of the operator+. #include <iostream> template<typename other_type> struct c { c(other_type v) : cs(v) {} struct subtype { subtype(other_type v) : val(v) {} other_type val; } cs; }; template<typename other_type> typename c<other_type>::subtype operator+(const typename c<other_type>::subtype& left, const typename c<other_type>::subtype& right) { return typename c<other_type>::subtype(left.val + right.val); } // This one works // c<a>::subtype operator+(const c<a>::subtype& left, // const c<a>::subtype& right) // { return c<a>::subtype(left.val + right.val); } int main() { c<int> c1 = 1; c<int> c2 = 2; c<int>::subtype cs3 = c1.cs + c2.cs; std::cerr << cs3.val << std::endl; } I think the reason is because the compiler (g++4.3) cannot guess the template type so it's searching for operator+<int> instead of operator+. What's the reason for that? What elegant solution can you suggest?

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  • C++0x Smart Pointer Comparisons: Inconsistent, what's the rationale?

    - by GManNickG
    In C++0x (n3126), smart pointers can be compared, both relationally and for equality. However, the way this is done seems inconsistent to me. For example, shared_ptr defines operator< be equivalent to: template <typename T, typename U> bool operator<(const shared_ptr<T>& a, const shared_ptr<T>& b) { return std::less<void*>()(a.get(), b.get()); } Using std::less provides total ordering with respect to pointer values, unlike a vanilla relational pointer comparison, which is unspecified. However, unique_ptr defines the same operator as: template <typename T1, typename D1, typename T2, typename D2> bool operator<(const unique_ptr<T1, D1>& a, const unique_ptr<T2, D2>& b) { return a.get() < b.get(); } It also defined the other relational operators in similar fashion. Why the change in method and "completeness"? That is, why does shared_ptr use std::less while unique_ptr uses the built-in operator<? And why doesn't shared_ptr also provide the other relational operators, like unique_ptr? I can understand the rationale behind either choice: with respect to method: it represents a pointer so just use the built-in pointer operators, versus it needs to be usable within an associative container so provide total ordering (like a vanilla pointer would get with the default std::less predicate template argument) with respect to completeness: it represents a pointer so provide all the same comparisons as a pointer, versus it is a class type and only needs to be less-than comparable to be used in an associative container, so only provide that requirement But I don't see why the choice changes depending on the smart pointer type. What am I missing? Bonus/related: std::shared_ptr seems to have followed from boost::shared_ptr, and the latter omits the other relational operators "by design" (and so std::shared_ptr does too). Why is this?

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  • C++11 Tidbits: Decltype (Part 2, trailing return type)

    - by Paolo Carlini
    Following on from last tidbit showing how the decltype operator essentially queries the type of an expression, the second part of this overview discusses how decltype can be syntactically combined with auto (itself the subject of the March 2010 tidbit). This combination can be used to specify trailing return types, also known informally as "late specified return types". Leaving aside the technical jargon, a simple example from section 8.3.5 of the C++11 standard usefully introduces this month's topic. Let's consider a template function like: template <class T, class U> ??? foo(T t, U u) { return t + u; } The question is: what should replace the question marks? The problem is that we are dealing with a template, thus we don't know at the outset the types of T and U. Even if they were restricted to be arithmetic builtin types, non-trivial rules in C++ relate the type of the sum to the types of T and U. In the past - in the GNU C++ runtime library too - programmers used to address these situations by way of rather ugly tricks involving __typeof__ which now, with decltype, could be rewritten as: template <class T, class U> decltype((*(T*)0) + (*(U*)0)) foo(T t, U u) { return t + u; } Of course the latter is guaranteed to work only for builtin arithmetic types, eg, '0' must make sense. In short: it's a hack. On the other hand, in C++11 you can use auto: template <class T, class U> auto foo(T t, U u) -> decltype(t + u) { return t + u; } This is much better. It's generic and a construct fully supported by the language. Finally, let's see a real-life example directly taken from the C++11 runtime library as implemented in GCC: template<typename _IteratorL, typename _IteratorR> inline auto operator-(const reverse_iterator<_IteratorL>& __x, const reverse_iterator<_IteratorR>& __y) -> decltype(__y.base() - __x.base()) { return __y.base() - __x.base(); } By now it should appear be completely straightforward. The availability of trailing return types in C++11 allowed fixing a real bug in the C++98 implementation of this operator (and many similar ones). In GCC, C++98 mode, this operator is: template<typename _IteratorL, typename _IteratorR> inline typename reverse_iterator<_IteratorL>::difference_type operator-(const reverse_iterator<_IteratorL>& __x, const reverse_iterator<_IteratorR>& __y) { return __y.base() - __x.base(); } This was guaranteed to work well with heterogeneous reverse_iterator types only if difference_type was the same for both types.

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  • Error in creating template class

    - by Luciano
    I found this vector template class implementation, but it doesn't compile on XCode. Header file: // File: myvector.h #ifndef _myvector_h #define _myvector_h template <typename ElemType> class MyVector { public: MyVector(); ~MyVector(); int size(); void add(ElemType s); ElemType getAt(int index); private: ElemType *arr; int numUsed, numAllocated; void doubleCapacity(); }; #include "myvector.cpp" #endif Implementation file: // File: myvector.cpp #include <iostream> #include "myvector.h" template <typename ElemType> MyVector<ElemType>::MyVector() { arr = new ElemType[2]; numAllocated = 2; numUsed = 0; } template <typename ElemType> MyVector<ElemType>::~MyVector() { delete[] arr; } template <typename ElemType> int MyVector<ElemType>::size() { return numUsed; } template <typename ElemType> ElemType MyVector<ElemType>::getAt(int index) { if (index < 0 || index >= size()) { std::cerr << "Out of Bounds"; abort(); } return arr[index]; } template <typename ElemType> void MyVector<ElemType>::add(ElemType s) { if (numUsed == numAllocated) doubleCapacity(); arr[numUsed++] = s; } template <typename ElemType> void MyVector<ElemType>::doubleCapacity() { ElemType *bigger = new ElemType[numAllocated*2]; for (int i = 0; i < numUsed; i++) bigger[i] = arr[i]; delete[] arr; arr = bigger; numAllocated*= 2; } If I try to compile as is, I get the following error: "Redefinition of 'MyVector::MyVector()'" The same error is displayed for every member function (.cpp file). In order to fix this, I removed the '#include "myvector.h"' on the .cpp file, but now I get a new error: "Expected constructor, destructor, or type conversion before '<' token". A similar error is displayed for every member as well. Interestingly enough, if I move all the .cpp code to the header file, it compiles fine. Does that mean I can't implement template classes in separate files?

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