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  • inspect C++ template instantiation

    - by aaa
    hello. Is there some utility which would allow me to inspect template instantiation? my compiler is g++ or Intel. Specific points I would like: Step by step instantiation. Instantiation backtrace (can hack this by crashing compiler. Better method?) Inspection of template parameters. Thanks

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  • C++: Constructor/destructor unresolved when not inline?

    - by Anamon
    In a plugin-based C++ project, I have a TmpClass that is used to exchange data between the main application and the plugins. Therefore the respective TmpClass.h is included in the abstract plugin interface class that is included by the main application project, and implemented by each plugin. As the plugins work on STL vectors of TmpClass instances, there needs to be a default constructor and destructor for the TmpClass. I had declared these in TmpClass.h: class TmpClass { TmpClass(); ~TmpClass(); } and implemented them in TmpClass.cpp. TmpClass::~TmpClass() {} TmpClass::TmpClass() {} However, when compiling plugins this leads to the linker complaining about two unresolved externals - the default constructor and destructor of TmpClass as required by the std::vector<TmpClass> template instantiation - even though all other functions I declare in TmpClass.h and implement in TmpClass.cpp work. As soon as I remove the (empty) default constructor and destructor from the .cpp file and inline them into the class declaration in the .h file, the plugins compile and work. Why is it that the default constructor and destructor have to be inline for this code to compile? Why does it even maatter? (I'm using MSVC++8).

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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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  • Typedef equivalence in function arguments

    - by Warren Seine
    Hi guys, The question is kind of hard to ask without an example so here it is: #include <vector> struct O { }; struct C { template <typename T> void function1(void (C::*callback)(const O*)); template <typename T> void function2(void (C::*callback)(const typename T::value_type)); void print(const O*); }; int main() { C c; c.function1< std::vector<O*> >(&C::print); // Success. c.function2< std::vector<O*> >(&C::print); // Fail. } The error that I am given is: error: no matching function for call to ‘C::function2(void (C::*)(const O*))’. Basically, the only difference between calls is that in function2, I'm more generic since I use the typedef std::vector<O*>::value_type which should resolve to O*, hence similar to function1. I'm using G++ 4.2.1 (I know it's old), but Comeau confirms I'm wrong. Why does the compilation fail?

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  • Why are there two implementations of std::sort (with and without a comparator) rather than one implementation with a default template parameter?

    - by PolyVox
    In my code I'm adopting a design strategy which is similar to some standard library algorithms in that the exact behavior can be customized by a function object. The simplest example is std::sort, where a function object can control how the comparison is made between objects. I notice that the Visual C++ provides two implementations of std::sort, which naturally involves code duplication. I would have imagined that it was instead possible to have only one implementation, and provide a default comparator (using operator< ) as a default template parameter. What is the rational behind two separate versions? Would my suggestion make the interface more complex in some way? Or result in confusing error messages when the object does not provide operator 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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  • 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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  • 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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  • Passing functor and function pointers interchangeably using a templated method in C++

    - by metroxylon
    I currently have a templated class, with a templated method. Works great with functors, but having trouble compiling for functions. Foo.h template <typename T> class Foo { public: // Constructor, destructor, etc... template <typename Func> void bar(T x, Func f); }; template <typename T> template <typename Func> Foo::bar(T x, Func f) { /* some code here */ } Main.cpp #include "Foo.h" template <typename T> class Functor { public: Functor() {} void operator()(T x) { /* ... */ } private: /* some attributes here */ }; void Function(T x) { /* ... */ } int main() { Foo<int> foo; foo.bar(2, Functor); // No problem foo.bar(2, Function); // <unresolved overloaded function type> return 0; }

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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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  • 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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  • 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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  • 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 can I derived from a templated/generic class based on that type in C# / C++

    - by stusmith
    Title probably doesn't make a lot of sense, so I'll start with some code: class Foo : public std::vector<Foo> { }; ... Foo f; f.push_back( Foo() ); Why is this allowed by the compiler? My brain is melting at this stage, so can anyone explain whether there are any reasons you would want to do this? Unfortunately I've just seen a similar pattern in some production C# code and wondered why anyone would use this pattern.

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  • C++ template overloading - wrong function called

    - by DeadMG
    template<typename T> T* Push(T* ptr); template<typename T> T* Push(T& ref); template<typename T, typename T1> T* Push(T1&& ref); I have int i = 0; Push<int>(i); But the compiler calls it ambiguous. How is that ambiguous? The second function is clearly the preferred match since it's more specialized. Especially since the T1&& won't bind to an lvalue unless I explicitly forward/move it. Sorry - i is an int. Otherwise, the question would make no sense, and I thought people would infer it since it's normally the loop iterator.

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  • Strange (atleast for me) behavior in Django template

    - by lud0h
    The following code snippet in a Django template (v 1.1) doesn't work. {{ item.vendors.all.0 }} == returns "Test" but the following code snippet, doesn't hide the paragraph! {% ifnotequal item.vendors.all.0 "Test" %} <p class="view_vendor">Vendor(s): {{item.vendors.all.0}} </p><br /> {% endifnotequal %} Any tips on what's wrong? Thanks.

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  • Strange overloading rules in C++

    - by bucels
    I'm trying to compile this code with GCC 4.5.0: #include <algorithm> #include <vector> template <typename T> void sort(T, T) {} int main() { std::vector<int> v; sort(v.begin(), v.end()); } But it doesn't seem to work: $ g++ -c nm.cpp nm.cpp: In function ‘int main()’: nm.cpp:9:28: error: call of overloaded ‘sort(std::vector<int>::iterator, std::vector<int>::iterator)’ is ambiguous nm.cpp:4:28: note: candidates are: void sort(T, T) [with T = __gnu_cxx::__normal_iterator<int*, std::vector<int> >] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/bits/stl_algo.h:5199:69: note: void std::sort(_RAIter, _RAIter) [with _RAIter = __gnu_cxx::__normal_iterator<int*, std::vector<int> >] Comeau compiles this code without errors. (4.3.10.1 Beta2, strict C++03, no C++0x) Is this valid C++?

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  • variadic constructors

    - by FredOverflow
    Are variadic constructors supposed to hide the implicitly generated ones, i.e. the default constructor and the copy constructor? struct Foo { template<typename... Args> Foo(Args&&... x) { std::cout << "inside the variadic constructor\n"; } }; int main() { Foo a; Foo b(a); } Somehow I was expecting this to print nothing after reading this answer, but it prints inside the variadic constructor twice on g++ 4.5.0 :( Is this behavior correct?

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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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  • 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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  • How to save link with tag e parameters in TextField

    - by xRobot
    I have this simple Post model: class Post(models.Model): title = models.CharField(_('title'), max_length=60, blank=True, null=True) body = models.TextField(_('body')) blog = models.ForeignKey(Blog, related_name="posts") user = models.ForeignKey(User) I want that when I insert in the form the links, the these links are saved in the body from this form: http://www.example.com or www.example.com to this form ( with tag and rel="nofollow" parameter ): <a href="http://www.example.com" rel="nofollow">www.example.com</a> How can I do this ? Thanks ^_^

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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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