boost lambda binding question

[baliga]

(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

[shunsuke]

On Jan 26, 9:50 am, [e-mail] 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

[baliga]

On Jan 25, 8:25 pm, "shunsuke" <[e-mail]:

On Jan 26, 9:50 am, [e-mail] 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?

[shunsuke]

"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, [e-mail] wrote:

On Jan 25, 8:25 pm, "shunsuke" <[e-mail]:

On Jan 26, 9:50 am, [e-mail] 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 -