Re: [Boost-users] [lambda] how to get rid of the reference argument in lambda functor?

AMDG Pobinger Gerhard Stefan wrote:

Hi *,

I was desparately trying to compile something like this (VS2008), always getting the error that it cannot convert from int* to int*&.

:boost::shared_ptr<int> c; ::boost::lambda::var(c) =::boost::lambda::constructor <::boost::shared_ptr<int>>() (::boost::lambda::new_ptr<int>()() )

I don't think that you want to evaluate new_ptr<int>() immediately. Try this (removing namespaces for clarity) (var(c) = constructor(new_ptr<int>()))()

[This is "short form" for the conventional code ::boost::shared_ptr<int>c=::boost::shared_ptr<int>(new int());] (1)

The reason is that all lambda functors take arguments by reference. In this case, make_const in front of shared_ptr did not solve the problem, it causes immediate evaluation of the new_ptr functor...

Is there any way to rewrite this (1) short line of code in boost-lambda?

In Christ, Steven Watanabe