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    <p>These line are simply adding the ability to handle Telegram objects to the standard output stream class.</p> <p>When you add a new class and you want output streams like <code>cout</code> to intelligently handle them, you need to add a new <code>&lt;&lt;</code> operator method which has the new object type as the second argument.</p> <p>What the code above is doing is exactly that. When you later execute the statements:</p> <pre><code>Telegram tg("Bob", "Hello, how are you?"); cout &lt;&lt; tg; </code></pre> <p>that function in your question will be called with the stream as the first argument and your <code>tg</code> object as the second argument and it will then be able to output the data in a format suitable to the class.</p> <p>This was actually one of the early C++ things I had trouble getting my head around. Although the class is supposed to be self-contained, you're actually adding something to a <em>different</em> class to handle the output. Once you understand why this is happening (because it's the <code>ostream</code> class that is responsible for outputting things rather than your own class), it will hopefully make sense.</p> <hr> <p><em>Hopefully</em> making it clearer with a simpler example:</p> <pre><code>1 inline std::ostream&amp; operator&lt;&lt;(std::ostream&amp; os, const Telegram&amp; t) { 2 os &lt;&lt; "message: " &lt;&lt; t.Msg; 3 return os; 4 } </code></pre> <p>Line 1 is simply the function definition. It allows you to return the stream itself (that you pass in) so you can chain <code>&lt;&lt;</code> segments. The <code>operator&lt;&lt;</code> is simply the function you're providing, which is the one called when you put <code>&lt;&lt; tg</code> into an output stream statement.</p> <p>Line 2 uses more basic <code>&lt;&lt;</code> statements that have already been defined (in this case, whatever type Msg is, probably a string).</p> <p>Then line 3 returns the stream, again to allow chaining of <code>&lt;&lt;</code> segments.</p> <p>The basic idea is to provide <code>operator&lt;&lt;</code> functions that build on existing <code>operator&lt;&lt;</code> function for the data types that constitute your type.</p> <hr> <p>And with a simple wrapper class containing just an <code>int</code>:</p> <pre><code>#include &lt;iostream&gt; // This is my simple class. class intWrapper { public: intWrapper (int x) { myInt = x; }; int getInt (void) { return myInt; } private: int myInt; // Must be friend to access private members. friend std::ostream&amp; operator&lt;&lt; (std::ostream&amp;, const intWrapper&amp;); }; // The actual output function. inline std::ostream&amp; operator&lt;&lt; (std::ostream&amp; os, const intWrapper&amp; t) { os &lt;&lt; "int: " &lt;&lt; t.myInt; return os; } // Main program for testing. // Output with getter and with ostream. int main (void) { class intWrapper x(7); std::cout &lt;&lt; x.getInt() &lt;&lt; std::endl; // ostream already knows about int. std::cout &lt;&lt; x &lt;&lt; std::endl; // And also intWrapper, due to the // function declared above. return 0; } </code></pre> <p>This outputs:</p> <pre><code>7 int: 7 </code></pre> <p>the first by just calling the getter function to retrieve the integer, the second by calling the <code>&lt;&lt;</code> operator function we added to <code>ostream</code>.</p>
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    1. POstd::ostream need help with function
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    1. COplease can you explain that with a simpler example and for a noob ??? sorry.
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      1. USjose gabriel
    2. CO`Telegram tg = new Telegram();`? Are you mixing up your languages, pax?
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      1. USBenjamin Lindley
    3. COand why theres another ostream object in the arguments???
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