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(std::cout << _1 << _2)("hello", 10);

This does not work. If 2 parameters are string, it works. If any one or
both parameters are integers, it does not work.

But If I change it to

boost::function<void(int, int)> f = std::cout << _1 << _2;
f(10, 10);

it works.

Question is why the first one does not work when one of the parameter
is integer.

Thanx,
-- baliga

http://baliga.blogdns.com/blog

On Jan 26, 9:50 am, bal...@gmail.com wrote:
> (std::cout << _1 << _2)("hello", 10);
>
> This does not work. If 2 parameters are string, it works. If any one or
> both parameters are integers, it does not work.
>
> But If I change it to
>
> boost::function<void(int, int)> f = std::cout << _1 << _2;
> f(10, 10);
>
> it works.
>
> Question is why the first one does not work when one of the parameter
> is integer.

Hi,

Lambda Functor can't accept a non-const rvalue(temporary).
See..
http://www.boost.org/doc/html/lambda...tual_arguments
http://std.dkuug.dk/jtc1/sc22/wg21/d...2002/n1385.htm

Thus,
(std::cout << _1 << _2)("hello", make_const(10));
works fine.

--
Shunsuke Sogame

On Jan 25, 8:25 pm, "shunsuke" <pstade...@gmail.com> wrote:
> On Jan 26, 9:50 am, bal...@gmail.com wrote:
>
> > (std::cout << _1 << _2)("hello", 10);
>
> > This does not work. If 2 parameters are string, it works. If any one or
> > both parameters are integers, it does not work.
>
> > But If I change it to
>
> > boost::function<void(int, int)> f = std::cout << _1 << _2;
> > f(10, 10);
>
> > it works.
>
> > Question is why the first one does not work when one of the parameter
> > is integer.Hi,
>
> Lambda Functor can't accept a non-const rvalue(temporary).
> See..http://www.boost.org/doc/html/lambda...2002/n1385.htm
>
> Thus,
> (std::cout << _1 << _2)("hello", make_const(10));
> works fine.
>
> --
> Shunsuke Sogame

Thanx for the solution. But I still have one question. why
make_const("hello") is not required for the string or is there any
implicit conversion going on within boost?

"reference to array" kicks in.
The reference seems passed to 'operator<<' as is.
Then, the 'operator<<' overload of 'char const*' is selected, AFAIK.


Regards,

--
Shunsuke Sogame


On Jan 27, 4:08 am, bal...@gmail.com wrote:
> On Jan 25, 8:25 pm, "shunsuke" <pstade...@gmail.com> wrote:
>
>
>
>
>
> > On Jan 26, 9:50 am, bal...@gmail.com wrote:
>
> > > (std::cout << _1 << _2)("hello", 10);
>
> > > This does not work. If 2 parameters are string, it works. If any one or
> > > both parameters are integers, it does not work.
>
> > > But If I change it to
>
> > > boost::function<void(int, int)> f = std::cout << _1 << _2;
> > > f(10, 10);
>
> > > it works.
>
> > > Question is why the first one does not work when one of the parameter
> > > is integer.Hi,
>
> > Lambda Functor can't accept a non-const rvalue(temporary).
> > See..http://www.boost.org/doc/html/lambda...#lambda.rvalue...
>
> > Thus,
> > (std::cout << _1 << _2)("hello", make_const(10));
> > works fine.
>
> > --
> > Shunsuke SogameThanx for the solution. But I still have one question. why
> make_const("hello") is not required for the string or is there any
> implicit conversion going on within boost?- Hide quoted text -- Show quoted text -