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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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  • Why can't I create a templated sublcass of System::Collections::Generic::IEnumerable<T>?

    - by fiirhok
    I want to create a generic IEnumerable implementation, to make it easier to wrap some native C++ classes. When I try to create the implementation using a template parameter as the parameter to IEnumerable, I get an error. Here's a simple version of what I came up with that demonstrates my problem: ref class A {}; template<class B> ref class Test : public System::Collections::Generic::IEnumerable<B^> // error C3225... {}; void test() { Test<A> ^a = gcnew Test<A>(); } On the indicated line, I get this error: error C3225: generic type argument for 'T' cannot be 'B ^', it must be a value type or a handle to a reference type If I use a different parent class, I don't see the problem: template<class P> ref class Parent {}; ref class A {}; template<class B> ref class Test : public Parent<B^> // no problem here {}; void test() { Test<A> ^a = gcnew Test<A>(); } I can work around it by adding another template parameter to the implementation type: ref class A {}; template<class B, class Enumerable> ref class Test : public Enumerable {}; void test() { using namespace System::Collections::Generic; Test<A, IEnumerable<A^>> ^a = gcnew Test<A, IEnumerable<A^>>(); } But this seems messy to me. Also, I'd just like to understand what's going on here - why doesn't the first way work?

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  • Is it possible to supply template parameters when calling operator()?

    - by Paul
    I'd like to use a template operator() but am not sure if it's possible. Here is a simple test case that won't compile. Is there something wrong with my syntax, or is this simply not possible? struct A { template<typename T> void f() { } template<typename T> void operator()() { } }; int main() { A a; a.f<int>(); // This compiles. a.operator()<int>(); // This compiles. a<int>(); // This won't compile. return 0; }

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  • C++ rvalue temporaries in template

    - by aaa
    hello. Can you please explain me the difference between mechanism of the following: int function(); template<class T> void function2(T&); void main() { function2(function()); // compiler error, instantiated as int & const int& v = function(); function2(v); // okay, instantiated as const int& } is my reasoning correct with respect to instantiation? why is not first instantiated as const T&? Thank you

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  • Obtain container type from (its) iterator type in C++ (STL)

    - by KRao
    It is easy given a container to get the associated iterators, example: std::vector<double>::iterator i; //An iterator to a std::vector<double> I was wondering if it is possible, given an iterator type, to deduce the type of the "corresponding container" (here I am assuming that for each container there is one and only one (non-const) iterator). More precisely, I would like a template metafunction that works with all STL containers (without having to specialize it manually for each single container) such that, for example: ContainerOf< std::vector<double>::iterator >::type evaluates to std::vector<double> Is it possible? If not, why? Thank you in advance for any help!

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  • How to know if the argument that is passed to the function is a class, union or enum in c++?

    - by Narek
    I want to define an operator<< for all enums, to cout the value and print that it is an enum like this: code: enum AnyEnum{A,B,C}; AnyEnum enm = A; cout << enm <<endl; output: This is an enum which has a value equal to 0 I know a way of doing this with Boost library by using is_enum struct. But I don’t understand how it works. So that's why, in general, I am interested how to identify if the veriable is a class type, union type or an enum (in compile time).

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  • How to split the definition of template friend funtion within template class?

    - by ~joke
    The following example compiles fine but I can't figure out how to separate declaration and definition of operator<<() is this particular case. Every time I try to split the definition friend is causing trouble and gcc complains the operator<<() definition must take exactly one argument. #include <iostream> template <typename T> class Test { public: Test(const T& value) : value_(value) {} template <typename STREAM> friend STREAM& operator<<(STREAM& os, const Test<T>& rhs) { os << rhs.value_; return os; } private: T value_; }; int main() { std::cout << Test<int>(5) << std::endl; }

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  • templated class : accessing derived normal-class methods

    - by user1019129
    I have something like this : class Container1 { public: method1() { ... } } class Container2 { public: method1() { ... } } template<class C = Container1> class X : public C { public: using C::method1(); ..... X(string& str) : C(str) {}; X& other_method() { method1(); ...; } } My question is why I have to use "using C::method1()", to be able to access the method.. Most of answers I found is for the case where templated-class inhering templated-class. Normally they mention using "this-", but this does not seem to work in this case. Can I do something else shorter... Also I'm suspecting the other error I'm getting is related to the same problem : no match call for (X<Container1>) (<std::string&>)

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  • 7 - drupal overriding theme fucntions gettting notices

    - by welovedesign
    So I am overriding a theme function by putting the contents in my template.php module, the problem is that it it throwing up loads of undefined index notices because there are lots of functions that are defined in the module. How can I define these in the template.php file and prevent the notices. Note: I know i can turn them off 'uc_cart_block_content' => array( 'variables' => array( 'help_text' => NULL, 'items' => NULL, 'item_count' => NULL, 'item_text' => NULL, 'total' => NULL, 'summary_links' => NULL, 'collapsed' => TRUE, ), 'file' => 'uc_cart.theme.inc', ),

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  • Template specialization to use default type if class member typedef does not exist

    - by Frank
    Hi Everyone, I'm trying to write code that uses a member typedef of a template argument, but want to supply a default type if the template argument does not have that typedef. A simplified example I've tried is this: struct DefaultType { DefaultType() { printf("Default "); } }; struct NonDefaultType { NonDefaultType() { printf("NonDefault "); } }; struct A {}; struct B { typedef NonDefaultType Type; }; template<typename T, typename Enable = void> struct Get_Type { typedef DefaultType Type; }; template<typename T> struct Get_Type< T, typename T::Type > { typedef typename T::Type Type; }; int main() { Get_Type::Type test1; Get_Type::Type test2; } I would expect this to print "Default NonDefault", but instead it prints "Default Default". My expectation is that the second line in main() should match the specialized version of Get_Type, because B::Type exists. However, this does not happen. Can anyone explain what's going on here and how to fix it, or another way to accomplish the same goal? Thank you.

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  • In a C++ template, is it allowed to return an object with specific type parameters?

    - by nieldw
    When I've got a template with certain type parameters, is it allowed for a function to return an object of this same template, but with different types? In other words, is the following allowed? template<class edgeDecor, class vertexDecor, bool dir> Graph<edgeDecor,int,dir> Graph<edgeDecor,vertexDecor,dir>::Dijkstra(vertex s, bool print = false) const { /* Construct new Graph with apropriate decorators */ Graph<edgeDecor,int,dir> span = new Graph<edgeDecor,int,dir>(); /* ... */ return span; }; If this is not allowed, how can I accomplish the same kind of thing?

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  • Optimal template for change content via XMLHTTPRequest with JQuery,PHP,SQL [closed]

    - by B.F.
    This is my method to handle XMLHTTPRequests. Avoids mysql request, foreign access, nerves user, double requests. jquery var allow=true; var is_loaded=""; $(document).ready(function(){ .... $(".xx").on("click",functio(){ if(allow){ allow=false; if(is_loaded!="that"){ $.post("job.php", {job:"that",word:"aaa",number:"123"},function(data){ $(".aaa").html(data); is_loaded="that"; }); } setTimeout(function(){allow=true},500); } .... }); job.php <?PHP ob_start('ob_gzhandler'); if(!isset($_SERVER['HTTP_X_REQUESTED_WITH']) or strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) != 'xmlhttprequest')exit("bad boy!"); if($_POST['job']=="that"){ include "includes/that.inc; } elseif($_POST['job']== .... ob_end_flush(); ?> that.inc if(!preg_match("/\w/",$_POST['word'])exit("bad boy!"); if(!is_numeric($_POST['number'])exit("bad boy!"); //exclude more. $path="temp/that_".$row['word']."txt"; if(file_exists($path) and filemtime("includes/that.inc")<$filemtime($path)){ readfile($path); } else{ include "includes/openSql.inc"; $call=sql_query("SELECT * FROM that WHERE name='".mysql_real_escape_string($_POST['word'])."'"); if(!$call)exit("ups"); $out=""; while($row=mysql_fetch_assoc($call)){ $out.=$_POST['word']." loves the color ".$row['color'].".<br/>"; } echo $out; $fn=fopen($path,"wb"); fputs($fn,$out); fclose($fn); } if something change at the database, you just have to delete involved files. Hope it was English.

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  • C++ Template Classes and Copy Construction

    - by themoondothshine
    Is there any way I can construct an new object from the given object if the template parameters of both objects are identical at run-time? For example: I have a template class with the declaration: template<typename _Type1, typename _Type2> class Object; Next, I have two instantiations of the template: template class Object<char, int>; template class Object<wchar_t, wint_t>; Now, I want to write a member function such as: template<typename _Type1, typename _Type2> Object<char, int> Object<_Type1, _Type2>::toCharObject() { if(__gnu_cxx::__are_same<_Type1, char>::__value) return *this; else { //Perform some kind of conversion and return an Object<char, int> } } I have tried a couple of techniques, such as using __gnu_cxx::__enable_if<__gnu_cxx::__are_same<_Type1, char>::__value, _Type1>::__type in a copy constructor for the Oject class, but I keep running into the error: error: conversion from ‘Object<wchar_t, wint_t>’ to non-scalar type ‘Object<char, int>’ requested Is there no way I can do this? Any help will be greatly appreciated!

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  • C++ template restrictions

    - by pingvinus
    I wondering is there any way to set restrictions on template class? Specify that every type substituted in template must have specific ancestor (realize some interface). template < class B > //and every B must be a child of abstract C class A { public: B * obj; int f() { return B::x + this->obj->f(); } }; Like = in haskell func :: (Ord a, Show b) => a -> b -> c

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  • Force type of C++ template

    - by gregseth
    Hi, I've a basic template class, but I'd like to restrain the type of the specialisation to a set of classes or types. e.g.: template <typename T> class MyClass { .../... private: T* _p; }; MyClass<std::string> a; // OK MYCLass<short> b; // OK MyClass<double> c; // not OK Those are just examples, the allowed types may vary. Is that even possible? If it is, how to do so? Thanks.

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  • C++ Template Usage

    - by MalcomTucker
    If I have a template definition like the one below, can someone provide a code sample for how I would actually instantiate an instance of this with two of my own classes? template <class T1, class T2> class LookUpTable { public: LookUpTable(); void set(Tl x, T2* y); T2* get(Tl x); }; Thanks.

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  • Template function overloading with identical signatures, why does this work?

    - by user1843978
    Minimal program: #include <stdio.h> #include <type_traits> template<typename S, typename T> int foo(typename T::type s) { return 1; } template<typename S, typename T> int foo(S s) { return 2; } int main(int argc, char* argv[]) { int x = 3; printf("%d\n", foo<int, std::enable_if<true, int>>(x)); return 0; } output: 1 Why doesn't this give a compile error? When the template code is generated, wouldn't the functions int foo(typename T::type search) and int foo(S& search) have the same signature? If you change the template function signatures a little bit, it still works (as I would expect given the example above): template<typename S, typename T> void foo(typename T::type s) { printf("a\n"); } template<typename S, typename T> void foo(S s) { printf("b\n"); } Yet this doesn't and yet the only difference is that one has an int signature and the other is defined by the first template parameter. template<typename T> void foo(typename T::type s) { printf("a\n"); } template<typename T> void foo(int s) { printf("b\n"); } I'm using code similar to this for a project I'm working on and I'm afraid that there's a subtly to the language that I'm not understanding that will cause some undefined behavior in certain cases. I should also mention that it does compile on both Clang and in VS11 so I don't think it's just a compiler bug.

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  • Checking if a function has C-linkage at compile-time [unsolvable]

    - by scjohnno
    Is there any way to check if a given function is declared with C-linkage (that is, with extern "C") at compile-time? I am developing a plugin system. Each plugin can supply factory functions to the plugin-loading code. However, this has to be done via name (and subsequent use of GetProcAddress or dlsym). This requires that the functions be declared with C-linkage so as to prevent name-mangling. It would be nice to be able to throw a compiler error if the referred-to function is declared with C++-linkage (as opposed to finding out at runtime when a function with that name does not exist). Here's a simplified example of what I mean: extern "C" void my_func() { } void my_other_func() { } // Replace this struct with one that actually works template<typename T> struct is_c_linkage { static const bool value = true; }; template<typename T> void assertCLinkage(T *func) { static_assert(is_c_linkage<T>::value, "Supplied function does not have C-linkage"); } int main() { assertCLinkage(my_func); // Should compile assertCLinkage(my_other_func); // Should NOT compile } Is there a possible implementation of is_c_linkage that would throw a compiler error for the second function, but not the first? I'm not sure that it's possible (though it may exist as a compiler extension, which I'd still like to know of). Thanks.

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