Re: [boost] Is it possible to get the argument's type of lambda_functor?

From: I Wei (i.c.code_at_[hidden])
Date: 2007-07-08 20:37:09

On 7/9/07, Peter Dimov <pdimov_at_[hidden]> wrote:
> I Wei wrote:
>
> > With the boost::function<>::argN_type I can get boost::function's
> > argument type. How can I do with the boost::lambda_functor?
>
> You can't because there is no such thing. A lambda functor such as _1 + 2
> can accept (almost) any argument x for which x+1 makes sense. What are you
> trying to do?
>
This is my situation:

#include "boost/function.hpp"
#include "boost/lambda/bind.hpp"
#include "boost/shared_ptr.hpp"

template <bool b>
struct getter
{
    template<typename T1, typename T2>
    void do_get(int n, T2 f)
    {
        // do something;
    }
};

template <>
struct getter<true>
{
    template<typename T1, typename T2>
    void do_get(int n, T2 f)
    {
        // do something else;
    }
};

template <typename T1, typename T2>
void func(int n, T2 f)
{
    typedef typename T2::arg1_type arg1_type;

    getter<
        boost::is_same<
                arg1_type,
                boost::shared_ptr<T1>
        >::value
    >do_get<T1, boost::function<void(arg1_type)> >(n, f);
}

class Foo
{
};

void foo(int n,  Foo t)
{
}

int main()
{
    func<Foo>(1, boost::lambda::bind(&foo, 0, boost::lambda::_1));

   return 0;
}

because the boost::lambda::bind return a lambda_functor,  the "
typedef typename T2::arg1_type arg1_type;" is wrong,  if I add the
second template argument to the caller func as func<Foo,
boost::function<void(Foo)> >, everything will be ok. But there are a
lot of similar calls with different second template argument, I want
the compiler deduce it by the caller function's actual argument. How
can I do that?