Re: Boost threads and overloaded call operator

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Mon, 22 Jun 2009 23:31:57 +0200
Message-ID:
<h1ot4j$qsu$1@news.eternal-september.org>
* pauldepstein@att.net:

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)); t.join
(); 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();


1.

     x();

2.

}


Cheers & hth.,

- Alf

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