Re: Boost threads and overloaded call operator

Mon, 22 Jun 2009 14:49:40 -0700 (PDT)
On Jun 22, 11:31 pm, "Alf P. Steinbach" <> wrote:


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

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
    void operator()()
         std::cout<<"I expected this line to occur once, not


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





Cheers & hth.,

- Alf

Due to hosting requirements I need visits to <url:


No ads, and there is some C++ stuff! :-) Just going there is good. Linkin=


to it is even better! Thanks in advance!

Sorry Alf, but I still don't understand.
If you omit the x(); line, I don't understand how the operator()() is
being called.
Please could you explain why the first three lines of code implement
the operator()()



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