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

From: Daniel Walker (daniel.j.walker_at_[hidden])
Date: 2007-07-09 11:59:10

On 7/8/07, I Wei <i.c.code_at_[hidden]> wrote:
> 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:
[snip]
>
> 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);
> }
[snip]
> 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?

I'm just guessing here (your example never actually invokes the lambda
functor), but I think you want to change func above to something
like...

template <typename T1, typename T2>
void func(T1 n, T2 f)
{
    typedef T1 arg1_type;
    // ...
}

... if you want to deduce the type of n.

I am not sure that you need to use boost::function at all. Also, this
seems like a very convoluted way of dispatching a function based on
whether or not it's called with a smart pointer. Why not overload
do_get for smart_pointer rather than specializing getter for the value
of is_same?

Daniel