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<float(float)> f = boost::lambda::bind(fx,2,boost::lambda::_1);

cheers
Arnaldur


"Chris Weed" <chrisweed@gmail.com>
Sent by: boost-users-bounces@lists.boost.org

19.10.2006 12:18
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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 <boost/function.hpp>
#include <boost/lambda/lambda.hpp>
#include <boost/lambda/bind.hpp>
#include <iostream>

float fx(float a, float b)
{
 std::cout << a << " " << b << std::endl;
 return a+b;
}

int main()
{
 boost::function<float(float)> 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 <arnaldur.gylfason@decode.is> wrote:
>
> a) You want to use _1 instead of _2
> b) _1/_2 are in namespace boost::lambda
> boost::function<float(float)> 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<float(float,float)> f = boost::lambda::bind(fx,boost::lambda::_2,2);
>    std::cout << f(100,3) << std::endl; // returns 5
>
> cheers
>
> Arnaldur
>
>
>
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