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