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    <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&lt; T >, foo&lt; 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 &lt;iostream&gt; using namespace std; template&lt; class T &gt; class foo { private: T _value; public: foo() : _value() {} template &lt;class U&gt; foo(const foo&lt;U&gt;&amp; 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&amp; that) : _value(that) {} friend const foo operator +(foo &amp;lhs,const foo &amp;rhs) { foo result(lhs._value+rhs._value); return result; }; friend const foo operator +(foo &amp;lhs,const T &amp;rhsval) { foo result(lhs._value+rhsval); return result; }; friend const foo operator +(const T &amp;lhsval,foo &amp;rhs) { foo result(lhsval+rhs._value); return result; }; friend foo&amp; operator +=(foo &amp;lhs,const foo &amp;rhs) { lhs._value+=rhs._value; return lhs; }; friend std::ostream&amp; operator&lt;&lt;(std::ostream&amp; out, const foo&amp; me){ return out &lt;&lt;me._value; } }; int main(){ foo&lt; int &gt; f, g; foo&lt; double &gt; dd; cout &lt;&lt;f&lt;&lt;endl; f = f + g; cout &lt;&lt;f&lt;&lt;endl; f += 3 ; cout &lt;&lt;f&lt;&lt;endl; f = f + 5; cout &lt;&lt;f&lt;&lt;endl; f = 7 + f; cout &lt;&lt;f&lt;&lt;endl; dd=dd+f; cout &lt;&lt;dd&lt;&lt;endl; dd=f+dd; cout &lt;&lt;dd&lt;&lt;endl; dd=dd+7.3; cout &lt;&lt;dd&lt;&lt;endl; } </code></pre>
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    1. COThis is exactly what I needed! Many thanks! I definitely need to grab myself a copy of Effective C++...
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    2. COgreat, I tried to write an answer as specific to the question as possible.
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    3. COWhat is the syntax for writing the friend functions out of line? If I remove the definition, I get an error "declares a non-template function".
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