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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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  • C++, what does this syntax mean?

    - by aaa
    i found this in this file: http://www.boost.org/doc/libs/1_43_0/boost/spirit/home/phoenix/core/actor.hpp What does this syntax means? struct actor ... { ... template <typename T0, typename T1> typename result<actor(T0&,T1&)>::type // this line thank you

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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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  • C++ Iterators and inheritance

    - by jomnis
    Have a quick question about what would be the best way to implement iterators in the following: Say I have a templated base class 'List' and two subclasses "ListImpl1" and "ListImpl2". The basic requirement of the base class is to be iterable i.e. I can do: for(List<T>::iterator it = list->begin(); it != list->end(); it++){ ... } I also want to allow iterator addition e.g.: for(List<T>::iterator it = list->begin()+5; it != list->end(); it++){ ... } So the problem is that the implementation of the iterator for ListImpl1 will be different to that for ListImpl2. I got around this by using a wrapper ListIterator containing a pointer to a ListIteratorImpl with subclasses ListIteratorImpl2 and ListIteratorImpl2, but it's all getting pretty messy, especially when you need to implement operator+ in the ListIterator. Any thoughts on a better design to get around these issues?

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  • What is causing this template-related compile error? (c++)

    - by Setien
    When I try to compile this: #include <map> #include <string> template <class T> class ZUniquePool { typedef std::map< int, T* > ZObjectMap; ZObjectMap m_objects; public: T * Get( int id ) { ZObjectMap::const_iterator it = m_objects.find( id ); if( it == m_objects.end() ) { T * p = new T; m_objects[ id ] = p; return p; } return m_objects[ id ]; } }; int main( int argc, char * args ) { ZUniquePool< std::string > pool; return 0; } I get this: main.cpp: In member function ‘T* ZUniquePool<T>::Get(int)’: main.cpp:12: error: expected `;' before ‘it’ main.cpp:13: error: ‘it’ was not declared in this scope I'm using GCC 4.2.1 on Mac OS X. It works in VS2008. I'm wondering whether it might be a variation of this problem: http://stackoverflow.com/questions/1364837/why-doesnt-this-c-template-code-compile But as my error output is only partially similar, and my code works in VS2008, I am not sure. Can anyone shed some light on what I am doing wrong?

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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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  • Problems Expanding an Array in C++

    - by dxq
    I'm writing a simulation for class, and part of it involves the reproduction of organisms. My organisms are kept in an array, and I need to increase the size of the array when they reproduce. Because I have multiple classes for multiple organisms, I used a template: template <class orgType> void expandarray(orgType* oldarray, int& numitems, int reproductioncount) { orgType *newarray = new orgType[numitems+reproductioncount]; for (int i=0; i<numitems; i++) { newarray[i] = oldarray[i]; } numitems += reproductioncount; delete[] oldarray; oldarray = newarray; newarray = NULL; } However, this template seems to be somehow corrupting my data. I can run the program fine without reproduction (commenting out the calls to expandarray), but calling this function causes my program to crash. The program does not crash DURING the expandarray function, but crashes on access violation later on. I've written functions to expand an array hundreds of times, and I have no idea what I screwed up this time. Is there something blatantly wrong in my function? Does it look right to you? EDIT: Thanks for everyone's help. I can't believe I missed something so obvious. In response to using std::vector: we haven't discussed it in class yet, and as silly as it seems, I need to write code using the methods we've been taught.

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  • template function error..

    - by sil3nt
    Hi there, I have function which takes in an parameter of a class called "Triple", and am returning the averge of 3 values of type float. template <typename ElemT> float average(Triple ElemT<float> &arg){ float pos1 = arg.getElem(1); float pos2 = arg.getElem(2); float pos3 = arg.getElem(3); return ( (pos1+pos2+po3) /3 ); } when i try compiling this i get q2b.cpp:32: error: template declaration of `float average' q2b.cpp:32: error: missing template arguments before "ElemT" not quite sure what this means.

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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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  • C++: How to require that one template type is derived from the other

    - by Will
    In a comparison operator: template<class R1, class R2> bool operator==(Manager<R1> m1, Manager<R2> m2) { return m1.internal_field == m2.internal_field; } Is there any way I could enforce that R1 and R2 must have a supertype or subtype relation? That is, I'd like to allow either R1 to be derived from R2, or R2 to be derived from R1, but disallow the comparison if R1 and R2 are unrelated types.

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  • C++ Template Question

    - by user323422
    see following code and please clear doubts1. as ABC is template why it not showing error when we put defination of ABC class member function in test.cpp 2.if i put test.cpp code in test.h , then it working fine // test.h template <typename T> class ABC { public: void foo( T& ); void bar( T& ); }; // test.cpp template <typename T> void ABC<T>::foo( T& ) {} // definition template <typename T> void ABC<T>::bar( T& ) {} // definition template void ABC<char>::foo( char & ); // 1 // main.cpp #include "test.h" int main() { ABC<char> a; a.foo(); // working a.bar(); // link error }

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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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  • 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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  • 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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