Error compile "no match for operator==" [duplicate] - c++

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Closed 10 years ago.
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no matching function error using template template parameters in a function
Good mornig I don't understand because I obtain compile error by implementing as global function the operator== (in class myList::iterator)
If I realize the operator== as member function the code compile.
The error is
error: no match for operator== in it == it1
note: candidate is:
note: tempalte<class T> bool operator==(const typename myList<T>::iterator&,const typename myList<T>::iterator&)
The code is:
#include <iostream>
#include <cstdlib>
#include <iterator>
#include <typeinfo>
template <class T>
class myList;
template <class T>
bool operator==(const typename myList<T>::iterator& lhs,const typename myList<T>::iterator& rhs);
template <class T>
bool operator!=(const typename myList<T>::iterator& lhs,const typename myList<T>::iterator& rhs);
template <class T>
class myList
{
private:
class myInfo;
public:
//CTR DEFAULT
myList():_pInfo(NULL)
{}
myList(T value):_pInfo(new myInfo(value))
{}
// class iterator;
// friend class iterator;
class iterator{
public:
//creo gli iteratori
iterator():_pMyInfoIt(NULL)
{}
iterator(myInfo* p):_pMyInfoIt(p)
{}
iterator(const iterator& it):_pMyInfoIt(it._pMyInfoIt)
{}
iterator& operator=(const iterator& it)
{
_pMyInfoIt = it._pMyInfoIt;
return *this;
}
//creo funzioni che lavorano sugli iteratori
/*
bool operator==(const iterator& rhs)
{
return _pMyInfoIt == rhs._pMyInfoIt;
}
*/
friend bool operator== <T>(const typename myList::iterator& lhs,const typename myList::iterator& rhs);
friend bool operator!= <T>(const typename myList<T>::iterator& lhs,const typename myList<T>::iterator& rhs);
private:
myInfo* _pMyInfoIt;
};
myList::iterator begin()
{
std::cout << __PRETTY_FUNCTION__ << std::endl;
return iterator(_pInfo);
}
private:
class myInfo
{
public:
myInfo(const T& data):_data(data),_pMyInfo(NULL)
{}
private:
T _data;
myInfo* _pMyInfo;
};
myInfo* _pInfo;
};
template <class T>
bool operator==(const typename myList<T>::iterator& lhs,const typename myList<T>::iterator& rhs)
{
std::cout << __PRETTY_FUNCTION__ << std::endl;
return lhs._pMyInfoIt == rhs._pMyInfoIt;
}
template <class T>
bool operator!=(const typename myList<T>::iterator& lhs,const typename myList<T>::iterator& rhs)
{
return !(lhs == rhs);
}
int main(int argc,char** argv){
myList<int> test;
myList<int>::iterator it = test.begin();
myList<int>::iterator it1 = test.begin();
std::cout << typeid(it1).name() << std::endl;
if(it == it1)
std::cout << "EQUAL" << std::endl;
return EXIT_SUCCESS;
}
thanks in advance

Answer 1

Compiler can not deduce T. Change to:
template <class IT, class=typename IT::iterator_category>
bool operator==(const IT& lhs, const IT& rhs)
2nd template parameter does simple SFINAE, so that only iterators will match.

Answer 2

Define the friend functions within the class body:
friend bool operator== <>(const typename myList<T>::iterator& lhs, const typename myList<T>::iterator& rhs)
{
std::cout << __LINE__ << std::endl;
return lhs._pMyInfoIt == rhs._pMyInfoIt;
}
friend bool operator!= <>(const typename myList<T>::iterator& lhs, const typename myList<T>::iterator& rhs)
{
return !(lhs == rhs);
}
It should work.

Answer 3

As pimanych says, declare the body of the friend function inside the class body:
friend bool operator==(const iterator& lhs, const iterator& rhs)
{
return lhs._pMyInfoIt == rhs._pMyInfoIt;
}
friend bool operator!=(const iterator& lhs, const iterator& rhs)
{
return !(lhs == rhs);
}
Note that these aren't really template functions, but are only declared when myList<T> (and myList<T>::iterator) is instantiated. You can check by attempting to compare a myList<int>::iterator and myList<float>::iterator - you should get a compiler error.
Look up the Barton-Nackman trick for more details.