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  • Specify a base classes template parameters while instantiating a derived class?

    - by DaClown
    Hi, I have no idea if the title makes any sense but I can't find the right words to descibe my "problem" in one line. Anyway, here is my problem. There is an interface for a search: template <typename InputType, typename ResultType> class Search { public: virtual void search (InputType) = 0; virtual void getResult(ResultType&) = 0; }; and several derived classes like: template <typename InputType, typename ResultType> class XMLSearch : public Search<InputType, ResultType> { public: void search (InputType) { ... }; void getResult(ResultType&) { ... }; }; The derived classes shall be used in the source code later on. I would like to hold a simple pointer to a Search without specifying the template parameters, then assign a new XMLSearch and thereby define the template parameters of Search and XMLSearch Search *s = new XMLSearch<int, int>(); I found a way that works syntactically like what I'm trying to do, but it seems a bit odd to really use it: template <typename T> class Derived; class Base { public: template <typename T> bool GetValue(T &value) { Derived<T> *castedThis=dynamic_cast<Derived<T>* >(this); if(castedThis) return castedThis->GetValue(value); return false; } virtual void Dummy() {} }; template <typename T> class Derived : public Base { public: Derived<T>() { mValue=17; } bool GetValue(T &value) { value=mValue; return true; } T mValue; }; int main(int argc, char* argv[]) { Base *v=new Derived<int>; int i=0; if(!v->GetValue(i)) std::cout<<"Wrong type int."<<std::endl; float f=0.0; if(!v->GetValue(f)) std::cout<<"Wrong type float."<<std::endl; std::cout<<i<<std::endl<<f; char c; std::cin>>c; return 0; } Is there a better way to accomplish this?

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  • Does template class/function specialization improves compilation/linker speed?

    - by Stormenet
    Suppose the following template class is heavily used in a project with mostly int as typename and linker speed is noticeably slower since the introduction of this class. template <typename T> class MyClass { void Print() { std::cout << m_tValue << std::endl;; } T m_tValue; } Will defining a class specialization benefit compilation speed? eg. void MyClass<int>::Print() { std::cout << m_tValue << std::endl; }

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  • Constant template parameter class manages to link externally

    - by the_drow
    I have a class foo with an enum template parameter and for some reason it links to two versions of the ctor in the cpp file. enum Enum { bar, baz }; template <Enum version = bar> class foo { public: foo(); }; // CPP File #include "foo.hpp" foo<bar>::foo() { cout << "bar"; } foo<baz>::foo() { cout << "baz"; } I'm using msvc 2008, is this the standard behavior? Are only type template parameters cannot be linked to cpp files?

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  • In Jeditable, how do I make it so that when I click the div to edit, the text box content has initial value that is processed?

    - by TIMEX
    When the user clicks on the div, jeditable will make a text box. However, I want the initial text to be done with function stripTags(), instead of what's on the page. The reason is that I'm using some URL techniques to turn plain text links into URLs. When the user clicks on the div, jeditable is turning them into <a href=>..</a> Is there a "beforeSubmit" option in jeditable? http://www.appelsiini.net/projects/jeditable

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  • C++ : Swapping template class elements of different types?

    - by metamemetics
    template< class T1, class T2 > class Pair { T1 first; T2 second; }; I'm being asked to write a swap() method so that the first element becomes the second and the second the first. I have: Pair<T2,T1> swap() { return Pair<T2,T1>(second, first); } But this returns a new object rather than swapping, where I think it needs to be a void method that changes its own data members. Is this possible to do since T1 and T2 are potentially different class types? In other words I can't simply set temp=first, first=second, second=temp because it would try to convert them to different types. I'm not sure why you would potentially want to have a template object that changes order of its types as it seems that would cause confusion but that appears to be what I'm being asked to do.

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  • std::basic_string full specialization (g++ conflict)

    - by SoapBox
    I am trying to define a full specialization of std::basic_string< char, char_traits<char>, allocator<char> > which is typedef'd (in g++) by the <string> header. The problem is, if I include <string> first, g++ sees the typedef as an instantiation of basic_string and gives me errors. If I do my specialization first then I have no issues. I should be able to define my specialization after <string> is included. What do I have to do to be able to do that? My Code: #include <bits/localefwd.h> //#include <string> // <- uncommenting this line causes compilation to fail namespace std { template<> class basic_string< char, char_traits<char>, allocator<char> > { public: int blah() { return 42; } size_t size() { return 0; } const char *c_str() { return ""; } void reserve(int) {} void clear() {} }; } #include <string> #include <iostream> int main() { std::cout << std::string().blah() << std::endl; } The above code works fine. But, if I uncomment the first #include <string> line, I get the following compiler errors: blah.cpp:7: error: specialization of ‘std::basic_string<char, std::char_traits<char>, std::allocator<char> >’ after instantiation blah.cpp:7: error: redefinition of ‘class std::basic_string<char, std::char_traits<char>, std::allocator<char> >’ /usr/include/c++/4.4/bits/stringfwd.h:52: error: previous definition of ‘class std::basic_string<char, std::char_traits<char>, std::allocator<char> >’ blah.cpp: In function ‘int main()’: blah.cpp:22: error: ‘class std::string’ has no member named ‘blah’ Line 52 of /usr/include/c++/4.4/bits/stringfwd.h: template<typename _CharT, typename _Traits = char_traits<_CharT>, typename _Alloc = allocator<_CharT> > class basic_string; As far as I know this is just a forward delcaration of the template, NOT an instantiation as g++ claims. Line 56 of /usr/include/c++/4.4/bits/stringfwd.h: typedef basic_string<char> string; As far as I know this is just a typedef, NOT an instantiation either. So why are these lines conflicting with my code? What can I do to fix this other than ensuring that my code is always included before <string>?

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  • C++ Types Impossible to Name

    - by Kirakun
    While reading Wikipedia's page on decltype, I was curious about the statement, Its [decltype's] primary intended use is in generic programming, where it is often difficult, or even impossible, to name types that depend on template parameters. While I can understand the difficulty part of that statement, what is an example where there is a need to name a type that cannot be named under C++03? EDIT: My point is that since everything in C++ has a declaration of types. Why would there ever be a case where it is impossible to name a type? Furthermore, aren't trait classes designed to yield type informations? Could trait classes be an alternative to decltype?

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  • static member specialization of templated child class and templated base class

    - by b3nj1
    I'm trying to have a templated class (here C) that inherits from another templated class (here A) and perform static member specialization (of int var here), but I cant get the right syntax to do so (if it's possible #include <iostream> template<typename derived> class A { public: static int var; }; //This one works fine class B :public A<B> { public: B() { std::cout << var << std::endl; } }; template<> int A<B>::var = 9; //This one doesn't works template<typename type> class C :public A<C<type> > { public: C() { std::cout << var << std::endl; } }; //template<> template<typename type> int A<C<type> >::a = 10; int main() { B b; C<int> c; return 0; } I put an example that works with a non templated class (here B) and i can get the static member specialization of var, but for C that just doesn't work. Here is what gcc tells me : test.cpp: In constructor ‘C<type>::C()’: test.cpp:29:26: error: ‘var’ was not declared in this scope test.cpp: At global scope: test.cpp:34:18: error: template definition of non-template ‘int A<C<type> >::a’ I'm using gcc version 4.6.3, thanks for any help

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  • What does "static" mean in the context of declaring global template functions?

    - by smf68
    I know what static means in the context of declaring global non-template functions (see e.g. What is a "static" function?), which is useful if you write a helper function in a header that is included from several different locations and want to avoid "duplicate definition" errors. So my question is: What does static mean in the context of declaring global template functions? Please note that I'm specifically asking about global, non-member template functions that do not belong to a class. In other words, what is the difference between the following two: template <typename T> void foo(T t) { /* implementation of foo here */ } template <typename T> static void bar(T t) { /* implementation of bar here */ }

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  • C++ template type deduction problem

    - by hamishmcn
    motivation: I would like to create a utility class so that instead of having to write: if( someVal == val1 || someVal == val2 || someVal == val3 ) I could instead write: if( is(someVal).in(val1, val2, val3) ) which is much closer to the mathematical 'a is an element of (b,c,d)' and also would save on a lot of typing when the variable name 'someVal' is long. Here is the code I have so far (for 2 and 3 values): template<class T> class is { private: T t_; public: is(T t) : t_(t) { } bool in(const T& v1, const T& v2) { return t_ == v1 || t_ == v2; } bool in(const T& v1, const T& v2, const T& v3) { return t_ == v1 || t_ == v2 || t_ == v3; } }; However it fails to compile if I write: is(1).in(3,4,5); instead I have to write is<int>(1).in(3,4,5); Which isn't too bad, but it would be better if somehow the compiler could figure out that the type is int with out me having to explicitly specify it. Is there anyway to do this or I am stuck with specifying it explicitly?

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  • Mapping integers to types using C++ template fails in a specific case

    - by Shailesh Kumar
    I am attempting to compile the following template based code in VC++ 2005. #include <iostream> using namespace std; /* * T is a template which maps an integer to a specific type. * The mapping happens through partial template specialization. * In the following T<1> is mapped to char, T<2> is mapped to long * and T<3> is mapped to float using partial template specializations */ template <int x> struct T { public: }; template<> struct T<1> { public: typedef char xType; }; template<> struct T<2> { public: typedef long xType; }; template<> struct T<3> { public: typedef float xType; }; // We can easily access the specific xType for a specific T<N> typedef T<3>::xType x3Type; /*! * In the following we are attempting to use T<N> inside another * template class T2<R> */ template<int r> struct T2 { //We can map T<r> to some other type T3 typedef T<r> T3; // The following line fails typedef T3::xType xType; }; int main() { T<1>::xType a1; cout << typeid(a1).name() << endl; T<2>::xType a2; cout << typeid(a2).name() << endl; T<3>::xType a3; cout << typeid(a3).name() << endl; return 0; } There is a particular line in the code which doesn't compile: typedef T3::xType xType; If I remove this line, compilation goes fine and the result is: char long float If I retain this line, compilation errors are observed. main.cpp(53) : warning C4346: 'T<x>::xType' : dependent name is not a type prefix with 'typename' to indicate a type main.cpp(54) : see reference to class template instantiation 'T2<r>' being compiled main.cpp(53) : error C2146: syntax error : missing ';' before identifier 'xType' main.cpp(53) : error C4430: missing type specifier - int assumed. Note: C++ does not support default-int I am not able to figure out how to make sure that T::xType can be treated as a type inside the T2 template. Any help is highly appreciated.

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  • template; Point<2, double>; Point<3, double>

    - by Oops
    Hi, I want to create my own Point struct it is only for purposes of learning C++. I have the following code: template <int dims, typename T> struct Point { T X[dims]; Point(){} Point( T X0, T X1 ) { X[0] = X0; X[1] = X1; } Point( T X0, T X1, T X2 ) { X[0] = X0; X[1] = X1; X[2] = X2; } Point<dims, int> toint() { //how to distinguish between 2D and 3D ??? Point<dims, int> ret = Point<dims, int>( (int)X[0], (int)X[1]); return ret; } std::string str(){ //how to distinguish between 2D and 3D ??? std::stringstream s; s << "{ X0: " << X[0] << " | X1: " << X[1] << " }"; return s.str(); } }; int main(void) { Point<2, double> p2d = Point<2, double>( 12.3, 45.6 ); Point<3, double> p3d = Point<3, double>( 12.3, 45.6, 78.9 ); Point<2, int> p2i = p2d.toint(); //OK Point<3, int> p3i = p3d.toint(); //m??? std::cout << p2d.str() << std::endl; //OK std::cout << p3d.str() << std::endl; //m??? std::cout << p2i.str() << std::endl; //m??? std::cout << p3i.str() << std::endl; //m??? char c; std::cin >> c; return 0; } of couse until now the output is not what I want. my questions is: how to take care of the dimensions of the Point (2D or 3D) in member functions of the Point? many thanks in advance Oops

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  • signature output operator overload

    - by coubeatczech
    hi, do you know, how to write signature of a function or method for operator<< for template class in C++? I want something like: template <class A class MyClass{ public: friend ostream & operator<<(ostream & os, MyClass<A mc); } ostream & operator<<(ostream & os, MyClass<A mc){ // some code return os; } But this just won't compile. Do anyone know, how to write it correctly?

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  • Why do I get the error "X is not a member of Y" even though X is a friend of Y?

    - by user1232138
    I am trying to write a binary tree. Why does the following code report error C2039, "'<<' : is not a member of 'btree<T'" even though the << operator has been declared as a friend function in the btree class? #include<iostream> using namespace std; template<class T> class btree { public: friend ostream& operator<<(ostream &,T); }; template<class T> ostream& btree<T>::operator<<(ostream &o,T s) { o<<s.i<<'\t'<<s.n; return o; }

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  • C++ creating generic template function specialisations

    - by Fire Lancer
    I know how to specialise a template function, however what I want to do here is specialise a function for all types which have a given method, eg: template<typename T> void foo(){...} template<typename T, if_exists(T::bar)>void foo(){...}//always use this one if the method T::bar exists T::bar in my classes is static and has different return types. I tried doing this by having an empty base class ("class HasBar{};") for my classes to derive from and using boost::enable_if with boost::is_base_of on my "specialised" version. However the problem then is that for classes that do have bar, the compiler cant resolve which one to use :(. template<typename T> typename boost::enable_if<boost::is_base_of(HasBar, T>, void>::type f() {...} I know that I could use boost::disable_if on the "normal" version, however I do not control the normal version (its provided by a third party library and its expected for specialisations to be made, I just don't really want to make explicit specialisations for my 20 or so classes), nor do I have that much control over the code using these functions, just the classes implementing T::bar and the function that uses it. Is there some way to tell the compiler to "always use this version if possible no matter what" without altering the other versions?

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  • Is the use of union in this matrix class completely safe?

    - by identitycrisisuk
    Unions aren't something I've used that often and after looking at a few other questions on them here it seems like there is almost always some kind of caveat where they might not work. Eg. structs possibly having unexpected padding or endian differences. Came across this in a math library I'm using though and I wondered if it is a totally safe usage. I assume that multidimensional arrays don't have any extra padding and since the type is the same for both definitions they are guaranteed to take up exactly the same amount of memory? template<typename T> class Matrix44T { ... union { T M[16]; T m[4][4]; } m; }; Are there any downsides to this setup? Would the order of definition make any difference to how this works?

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  • Adapting non-iterable containers to be iterated via custom templatized iterator

    - by DAldridge
    I have some classes, which for various reasons out of scope of this discussion, I cannot modify (irrelevant implementation details omitted): class Foo { /* ... irrelevant public interface ... */ }; class Bar { public: Foo& get_foo(size_t index) { /* whatever */ } size_t size_foo() { /* whatever */ } }; (There are many similar 'Foo' and 'Bar' classes I'm dealing with, and it's all generated code from elsewhere and stuff I don't want to subclass, etc.) [Edit: clarification - although there are many similar 'Foo' and 'Bar' classes, it is guaranteed that each "outer" class will have the getter and size methods. Only the getter method name and return type will differ for each "outer", based on whatever it's "inner" contained type is. So, if I have Baz which contains Quux instances, there will be Quux& Baz::get_quux(size_t index), and size_t Baz::size_quux().] Given the design of the Bar class, you cannot easily use it in STL algorithms (e.g. for_each, find_if, etc.), and must do imperative loops rather than taking a functional approach (reasons why I prefer the latter is also out of scope for this discussion): Bar b; size_t numFoo = b.size_foo(); for (int fooIdx = 0; fooIdx < numFoo; ++fooIdx) { Foo& f = b.get_foo(fooIdx); /* ... do stuff with 'f' ... */ } So... I've never created a custom iterator, and after reading various questions/answers on S.O. about iterator_traits and the like, I came up with this (currently half-baked) "solution": First, the custom iterator mechanism (NOTE: all uses of 'function' and 'bind' are from std::tr1 in MSVC9): // Iterator mechanism... template <typename TOuter, typename TInner> class ContainerIterator : public std::iterator<std::input_iterator_tag, TInner> { public: typedef function<TInner& (size_t)> func_type; ContainerIterator(const ContainerIterator& other) : mFunc(other.mFunc), mIndex(other.mIndex) {} ContainerIterator& operator++() { ++mIndex; return *this; } bool operator==(const ContainerIterator& other) { return ((mFunc.target<TOuter>() == other.mFunc.target<TOuter>()) && (mIndex == other.mIndex)); } bool operator!=(const ContainerIterator& other) { return !(*this == other); } TInner& operator*() { return mFunc(mIndex); } private: template<typename TOuter, typename TInner> friend class ContainerProxy; ContainerIterator(func_type func, size_t index = 0) : mFunc(func), mIndex(index) {} function<TInner& (size_t)> mFunc; size_t mIndex; }; Next, the mechanism by which I get valid iterators representing begin and end of the inner container: // Proxy(?) to the outer class instance, providing a way to get begin() and end() // iterators to the inner contained instances... template <typename TOuter, typename TInner> class ContainerProxy { public: typedef function<TInner& (size_t)> access_func_type; typedef function<size_t ()> size_func_type; typedef ContainerIterator<TOuter, TInner> iter_type; ContainerProxy(access_func_type accessFunc, size_func_type sizeFunc) : mAccessFunc(accessFunc), mSizeFunc(sizeFunc) {} iter_type begin() const { size_t numItems = mSizeFunc(); if (0 == numItems) return end(); else return ContainerIterator<TOuter, TInner>(mAccessFunc, 0); } iter_type end() const { size_t numItems = mSizeFunc(); return ContainerIterator<TOuter, TInner>(mAccessFunc, numItems); } private: access_func_type mAccessFunc; size_func_type mSizeFunc; }; I can use these classes in the following manner: // Sample function object for taking action on an LMX inner class instance yielded // by iteration... template <typename TInner> class SomeTInnerFunctor { public: void operator()(const TInner& inner) { /* ... whatever ... */ } }; // Example of iterating over an outer class instance's inner container... Bar b; /* assume populated which contained items ... */ ContainerProxy<Bar, Foo> bProxy( bind(&Bar::get_foo, b, _1), bind(&Bar::size_foo, b)); for_each(bProxy.begin(), bProxy.end(), SomeTInnerFunctor<Foo>()); Empirically, this solution functions correctly (minus any copy/paste or typos I may have introduced when editing the above for brevity). So, finally, the actual question: I don't like requiring the use of bind() and _1 placeholders, etcetera by the caller. All they really care about is: outer type, inner type, outer type's method to fetch inner instances, outer type's method to fetch count inner instances. Is there any way to "hide" the bind in the body of the template classes somehow? I've been unable to find a way to separately supply template parameters for the types and inner methods separately... Thanks! David

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  • Is there a Visual Studio (or freeware) equivalent for Expression Blend's "Edit Template" feature?

    - by DanM
    In Expression Blend, you can view and edit the control template of objects in the "Objects and Timeline" panel. I'm wondering if there's an equivalent feature in Visual Studio or if there's something free (or very inexpensive) I can download that will allow me to do this. Here's a screen cap from Expression Blend that shows what I'm talking about: Doing this for DataGrid results in the following: <Style x:Key="DataGridStyle1" TargetType="{x:Type Custom:DataGrid}"> ... <Setter Property="Template"> <Setter.Value> <ControlTemplate TargetType="{x:Type Custom:DataGrid}"> ... </ControlTemplate> </Setter.Value> </Setter> <Style.Triggers> <Trigger Property="IsGrouping" Value="True"> <Setter Property="ScrollViewer.CanContentScroll" Value="False"/> </Trigger> </Style.Triggers> </Style> (The ... is of course replaced with setters and the contents of the control template.) This is a very useful starting point if you want to create a custom style and template for a control. It seems like you can do pretty much anything you can do in Blend in Studio, but this one is eluding me. Any ideas? Edit I'm also curious if this feature will be in Visual Studio 2010. Anyone know?

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  • Is call to function object inlined?

    - by dehmann
    In the following code, Foo::add calls a function via a function object: struct Plus { inline int operator()(int x, int y) const { return x + y; } }; template<class Fct> struct Foo { Fct fct; Foo(Fct f) : fct(f) {} inline int add(int x, int y) { return fct(x,y); // same efficiency adding directly? } }; Is this the same efficiency as calling x+y directly in Foo::add? In other words, does the compiler typically directly replace fct(x,y) with the actual call, inlining the code, when compiling with optimizations enabled?

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  • C++ CRTP question

    - by aaa
    following piece of code does not compile, the problem is in T::rank not be inaccessible (I think) or uninitialized in parent template. Can you tell me exactly what the problem is? is passing rank explicitly the only way? or is there a way to query tensor class directly? Thank you #include <boost/utility/enable_if.hpp> template<class T, // size_t N, class enable = void> struct tensor_operator; // template<class T, size_t N> template<class T> struct tensor_operator<T, typename boost::enable_if_c< T::rank == 4>::type > { tensor_operator(T &tensor) : tensor_(tensor) {} T& operator()(int i,int j,int k,int l) { return tensor_.layout.element_at(i, j, k, l); } T &tensor_; }; template<size_t N, typename T = double> // struct tensor : tensor_operator<tensor<N,T>, N> { struct tensor : tensor_operator<tensor<N,T> > { static const size_t rank = N; }; I know the workaround, however am interested in mechanics of template instantiation for self-education

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  • C++ Generic List Assignment

    - by S73417H
    I've clearly been stuck in Java land for too long... Is it possible to do the C++ equivalent of the following Java code: // Method List<Bar> getBars() { return new LinkedList<Bar>(); } // Assignment statement. List<Foo> stuff = getBars(); Where Foo is a sub-class of Bar. So in C++.... std::list<Bar> & getBars() { std::list<Bar> bars; return bars; } std::list<Foo> stuff = getBars(); Hope that makes sense....

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  • Shortcut to get all private field names combined with && operator as a generated line of code?

    - by burak ozdogan
    Hi, Is it possible to use reflection when we write a code-template? I was just thinking if it is possible to write a piece of code-template that will return me a list of all private field in the class separated by && operator in Visual Stuudio? Let's say, I will enter "getAllPrivates&&" (or press a shortcut key) in the VS editor and it will return me something like this: private bool _privateFiel1 = false; private bool _privateFiel2 = true; private bool _privateFiel3 = false; private bool _privateFiel4 = false; private bool _privateFiel5 = true; // Here I press the shortcut or call the code-template and get his: _privateFiel1 && _privateFiel2 && _privateFiel3 && _privateFiel4 && _privateFiel5

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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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