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    <p>The simplest thing that would work:</p> <pre><code>string operator + (const Example&lt;char&gt; &amp;a, const Example&lt;char&gt; &amp;b) { return { a.data, b.data }; } </code></pre> <p>If you have an 'older' compiler:</p> <pre><code>string operator + (const Example&lt;char&gt; &amp;a, const Example&lt;char&gt; &amp;b) { char both[] = {a.data, b.data}; return string(both, both+2); } </code></pre> <p>See it live: <a href="http://liveworkspace.org/code/2ORW8E%240" rel="nofollow">http://liveworkspace.org/code/2ORW8E$0</a></p> <p><strong>UPDATE</strong> To the edit question: There is also a big problem (infinite recursion) here:</p> <pre><code>template &lt;class U&gt; U operator + (const Example&lt;U&gt; &amp;a, const Example&lt;U&gt; &amp;b) { U c; c = a+b; // NEEDS TO BE a.data + b.data; return(c); } </code></pre> <p>Here is a fixed version:</p> <pre><code>#include &lt;iostream&gt; #include &lt;string&gt; using namespace std; template &lt;class T&gt; class Example { private: T data; public: Example() : data() { } void setData(T elem) { data = elem; } template &lt;class U&gt; friend ostream&amp; operator &lt;&lt; (ostream &amp;, const Example&lt;U&gt;&amp;); friend ostream&amp; operator &lt;&lt; (ostream &amp;, const Example&lt;char&gt;&amp;); friend string operator + (const Example&lt;char&gt; &amp;, const Example&lt;char&gt; &amp;); template &lt;class U&gt; friend U operator + (const Example&lt;U&gt; &amp;, const Example&lt;U&gt; &amp;); }; template &lt;class U&gt; U operator + (const Example&lt;U&gt; &amp;a, const Example&lt;U&gt; &amp;b) { U c; c = a.data+b.data; return(c); } string operator + (const Example&lt;char&gt; &amp;a, const Example&lt;char&gt; &amp;b) { char both[] = {a.data, b.data}; return string(both, both+2); } template &lt;class T&gt; ostream&amp; operator &lt;&lt; (ostream &amp;o, const Example&lt;T&gt; &amp;t) { o &lt;&lt; t.data; return o; } ostream&amp; operator &lt;&lt; (ostream &amp;o, const Example&lt;char&gt; &amp;t) { o &lt;&lt; "'" &lt;&lt; t.data &lt;&lt; "'"; return o; } int main() { Example&lt;int&gt; tInt1, tInt2; Example&lt;char&gt; tChar1, tChar2; tInt1.setData(15); tInt2.setData(30); tChar1.setData('A'); tChar2.setData('B'); cout &lt;&lt; tInt1 &lt;&lt; " + " &lt;&lt; tInt2 &lt;&lt; " = " &lt;&lt; (tInt1 + tInt2) &lt;&lt; endl; cout &lt;&lt; tChar1 &lt;&lt; " + " &lt;&lt; tChar2 &lt;&lt; " = " &lt;&lt; (tChar1 + tChar2) &lt;&lt; endl; return 0; } </code></pre>
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    1. POConverting two characters passed by reference into a string?
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    1. COthe op's class doth not haz member `data`, and visual c++ doth not support braced initializer sintax, besides i'm not sure that it would realy denot an array in this case
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      1. USCheers and hth. - Alf
    2. CO@Cheersandhth.-Alf thanks for pointing that out. Of course, everyone hath made assumptions - I posted mine so people can judge my assumptions for themselves :)
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    3. CO@Cheersandhth.-Alf ironically, the OP's class _doth_ have `data` after all, and a few more issues which I touched upon in my updated answer :)
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