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    <pre><code>operator std::map&lt;T, U&gt;() { return m_map; } </code></pre> <p>This is user-defined <em>conversion</em> function.</p> <p>That means, you can write this:</p> <pre><code>//obj is an object of type create_map create_map&lt;int,std::string&gt; obj(1,"Nawaz"); //obj implicitly converts into std::map type! std::map&lt;int,std::string&gt; map_inst=obj; </code></pre> <hr> <p>See this topic to know more about <strong>user-defined conversion function</strong>:</p> <p><a href="https://stackoverflow.com/questions/3008690/user-defined-conversions-in-c">User Defined Conversions in C++</a></p> <p>You can see this as well: <a href="http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/index.jsp?topic=/com.ibm.xlcpp8a.doc/language/ref/implicit_conversion_sequences.htm" rel="nofollow noreferrer">Implicit conversion sequences (C++ only)</a></p> <hr> <pre><code>create_map&lt;T, U&gt;&amp; operator()(const T&amp; key, const U&amp; val) { m_map[key] = val; return *this; } </code></pre> <p>This actually overloads the <code>operator()</code>, which internally inserts or updates (key,val) pair into the <code>m_map</code>; just see the function definition as to what it does. </p> <p>So using this function you can write code like this,</p> <pre><code>obj(2,"Sarfaraz"); //this inserts the pair into the underlying m_map; </code></pre> <p>I would also suggest you to explore <code>std::map</code> a bit more, especially the overloaded <code>operator[]</code> in <code>std::map</code>.</p>
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    1. POI need someone who explain me these lines of code
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    1. COsorry i really dont understandthat. i really need an explanation line by line, is really important thank you.
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    2. COso it just convert the create_map object to a map object when necessary? and can you explain me please the other operator overloading? and how is it used? im sorry im kind of noob. thank you very much.
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    3. CO@jose: Exactly.
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