[function][lambda] Create a boost function to a bind functor

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Chris Weed

19 Oct 2006 19 Oct '06

5:03 a.m.

Hi, I would like to create a boost::function to a lambda::bind functor such as float fx(float a, float b) { return a+b; } int main() { boost::function f = boost::lambda::bind(fx,_2,2); std::cout << f(3) << std::endl; return 0; } I would expect this to bind 2 to the "b" argument of fx, and print "5". Instead with msvc-8.0 I get a bunch or compile errors. Any help would be greatly appreciated. Chris

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Arnaldur Gylfason

19 Oct 19 Oct

11:54 a.m.

New subject: [function][lambda] Create a boost function to a bind functor

a) You want to use _1 instead of _2 b) _1/_2 are in namespace boost::lambda boost::function f = boost::lambda::bind(fx,boost::lambda::_1,2); It is actually possible to use _2 but then the functor takes to args and ignores the first: boost::function f = boost::lambda::bind(fx,boost::lambda::_2,2); std::cout << f(100,3) << std::endl; // returns 5 cheers Arnaldur "Chris Weed" Sent by: boost-users-bounces@lists.boost.org 19.10.2006 05:04 Please respond to boost-users@lists.boost.org To boost-users@lists.boost.org cc Subject [Boost-users] [function][lambda] Create a boost function to a bind functor Hi, I would like to create a boost::function to a lambda::bind functor such as float fx(float a, float b) { return a+b; } int main() { boost::function f = boost::lambda::bind(fx,_2,2); std::cout << f(3) << std::endl; return 0; } I would expect this to bind 2 to the "b" argument of fx, and print "5". Instead with msvc-8.0 I get a bunch or compile errors. Any help would be greatly appreciated. Chris _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users

Chris Weed

12:17 p.m.

New subject: [function][lambda] Create a boost function to a bind functor

Hi, I can get the following code to compile and it prints 3 2 5 but I don't quite understand why it doesn't print 2 3 5 or how to get it to do so. I thought since it is binding 2 to _1, this would bind it to the a argument. #include #include #include #include <iostream> float fx(float a, float b) { std::cout << a << " " << b << std::endl; return a+b; } int main() { boost::function f = boost::lambda::bind(fx,boost::lambda::_1,2); std::cout << f(3) << std::endl; return 0; } Thanks, Chris On 10/19/06, Arnaldur Gylfason wrote:

...

a) You want to use _1 instead of _2 b) _1/_2 are in namespace boost::lambda boost::function f = boost::lambda::bind(fx,boost::lambda::_1,2);

It is actually possible to use _2 but then the functor takes to args and ignores the first: boost::function f = boost::lambda::bind(fx,boost::lambda::_2,2); std::cout << f(100,3) << std::endl; // returns 5

cheers

Arnaldur

Peter Dimov

12:27 p.m.

New subject: [function][lambda] Create a boost function to abind functor

Chris Weed wrote:

...

Hi, I can get the following code to compile and it prints 3 2 5 but I don't quite understand why it doesn't print 2 3 5 or how to get it to do so. I thought since it is binding 2 to _1, this would bind it to the a argument.

Calling bind( fx, _1, 2 ) with 3 will put the '3' in the _1 and pass the 2 intact, the result is fx( 3, 2 ). bind( fx, _1, _1 ) will call fx( 3, 3 ). bind( fx, 2, _1 ) calls fx( 2, 3 ). bind( fx, 4, 5 ) calls fx( 4, 5 ). Does it make sense now? bind is not an interpreter with a mini-language ("bind the second argument to 2"), you just pass the argument values, and _1 is substituted with the first input argument (3 in your case).

Chris Weed

12:59 p.m.

New subject: [function][lambda] Create a boost function to abind functor

Yes, That clears up my understanding. Thanks, Chris On 10/19/06, Peter Dimov wrote:

...

Chris Weed wrote:

...
Hi, I can get the following code to compile and it prints 3 2 5 but I don't quite understand why it doesn't print 2 3 5 or how to get it to do so. I thought since it is binding 2 to _1, this would bind it to the a argument.

Calling bind( fx, _1, 2 ) with 3 will put the '3' in the _1 and pass the 2 intact, the result is fx( 3, 2 ). bind( fx, _1, _1 ) will call fx( 3, 3 ). bind( fx, 2, _1 ) calls fx( 2, 3 ). bind( fx, 4, 5 ) calls fx( 4, 5 ).

Does it make sense now? bind is not an interpreter with a mini-language ("bind the second argument to 2"), you just pass the argument values, and _1 is substituted with the first input argument (3 in your case).

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Arnaldur Gylfason

12:32 p.m.

New subject: [function][lambda] Create a boost function to a bind functor

It is binding 2 to the second argument of the function (which is b) but _1 to a which means the argument you call f with is passed as a to your function. Hence f(3) prints 3 2 and returns 5. If you want 2 3 you need: boost::function f = boost::lambda::bind(fx,2,boost::lambda::_1); cheers Arnaldur "Chris Weed" Sent by: boost-users-bounces@lists.boost.org 19.10.2006 12:18 Please respond to boost-users@lists.boost.org To boost-users@lists.boost.org cc Subject Re: [Boost-users] [function][lambda] Create a boost function to a bind functor Hi, I can get the following code to compile and it prints 3 2 5 but I don't quite understand why it doesn't print 2 3 5 or how to get it to do so. I thought since it is binding 2 to _1, this would bind it to the a argument. #include #include #include #include <iostream> float fx(float a, float b) { std::cout << a << " " << b << std::endl; return a+b; } int main() { boost::function f = boost::lambda::bind(fx,boost::lambda::_1,2); std::cout << f(3) << std::endl; return 0; } Thanks, Chris On 10/19/06, Arnaldur Gylfason wrote:

...

a) You want to use _1 instead of _2 b) _1/_2 are in namespace boost::lambda boost::function f =

boost::lambda::bind(fx,boost::lambda::_1,2);

...

It is actually possible to use _2 but then the functor takes to args and

ignores the first:

...

boost::function f = boost::lambda::bind(fx,boost::lambda::_2,2); std::cout << f(100,3) << std::endl; // returns 5

cheers

Arnaldur

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Arnaldur Gylfason
Chris Weed
Peter Dimov