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copied!<p>The reason it does not <em>just work</em> is that implicit type conversions (that is, via constructors) do not apply during template argument deduction. But it works if you make the outside operator a friend since then the type T is know, allowing the compiler to investigate what can be casted to make the arguments match.</p> <p>I made an example based on yours (but removed C++11 stuff), inspired by Item 46 (a rational number class) in Scott Meyers Effective C++ (ed 3). Your question is almost an exact match to that item. Scott also notes that ... "this use of friend is not related to the access of non-public parts of the class."</p> <p>This will also allow work with mixes of foo< T >, foo< U > etc as long as T and U can be added etc.</p> <p>Also look at this post: <a href="https://stackoverflow.com/questions/2613645/c-addition-overload-ambiguity">C++ addition overload ambiguity</a></p> <pre><code>#include <iostream> using namespace std; template< class T > class foo { private: T _value; public: foo() : _value() {} template <class U> foo(const foo<U>& that) : _value(that.getval()) {} // I'm sure this it can be done without this being public also; T getval() const { return _value ; }; foo(const T& that) : _value(that) {} friend const foo operator +(foo &lhs,const foo &rhs) { foo result(lhs._value+rhs._value); return result; }; friend const foo operator +(foo &lhs,const T &rhsval) { foo result(lhs._value+rhsval); return result; }; friend const foo operator +(const T &lhsval,foo &rhs) { foo result(lhsval+rhs._value); return result; }; friend foo& operator +=(foo &lhs,const foo &rhs) { lhs._value+=rhs._value; return lhs; }; friend std::ostream& operator<<(std::ostream& out, const foo& me){ return out <<me._value; } }; int main(){ foo< int > f, g; foo< double > dd; cout <<f<<endl; f = f + g; cout <<f<<endl; f += 3 ; cout <<f<<endl; f = f + 5; cout <<f<<endl; f = 7 + f; cout <<f<<endl; dd=dd+f; cout <<dd<<endl; dd=f+dd; cout <<dd<<endl; dd=dd+7.3; cout <<dd<<endl; } </code></pre>

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