Re: Boost threads and overloaded call operator

I ran the following code which outputs two lines of text on the
screen. I can understand why the line x(); outputs a line of text
because that's the normal form of the overloaded call operator
operator()()

But it's puzzling to me why the lines boost::thread t((x)); =

(); also output a line of text on the screen.
To me the line boost::thread t((x)) doesn't seem to be applying the
overloaded call operator.
I would have thought that the overloaded call operator is implemented
via x(); or via x.operator();

I'd be grateful if someone could explain why boost::thread t((x));
apparently implements the overloaded call operator.

Thank you very much,

Paul Epstein

#include <boost\thread\thread.hpp>
#include <iostream>

class SayHello
{
public:
    void operator()()
    {
         std::cout<<"I expected this line to occur once, not
twice!"<<std::endl;
    }

};

int main()
{ SayHello x;
    boost::thread t((x));
    t.join();

   x();

}