can copy constructor be a conversion operator?
<p>Consider the following code.</p>

<pre><code>#include &lt;iostream&gt;

using namespace std;


class Test
{
        private:
                int x,y;
        public:
                Test () {
                        cout &lt;&lt;" Inside Constructor "&lt;&lt;endl;
                        x=100;
                }   
                explicit Test (const Test &amp; t)
                {   
                        cout &lt;&lt;"Inside Copy Constructor "&lt;&lt;endl;
                        x = t.x;
                }   
                void display()
                {   
                        cout &lt;&lt;" X is "&lt;&lt;x&lt;&lt;endl;
                }   

};



int main (int argc, char ** argv){
  Test t;
  t.display(); 

 cout &lt;&lt;"--- Using Copy constructor "&lt;&lt;endl;
 Test t2(t);
 t2.display (); 

 Test t3=t2;
 t3.display (); 

}
</code></pre>

<p>Test (const Test &amp; t)  -> is a copy constructor</p>

<p><strong>Question:</strong></p>

<p>Is the same used as a "Conversion Operator" ?
Test t3  = t2     [ Here copy Constructor is treated as a conversion operator]</p>

<p>I am not sure if my understanding is correct?. Kindly correct me if i am wrong? </p>
