Re: [Boost-users] naming delayed variables string in boost lamdba

Subject: Re: [Boost-users] naming delayed variables string in boost lamdba
From: Ryan McConnehey (mccorywork_at_[hidden])
Date: 2009-11-11 12:46:54

OvermindDL1 wrote:
> On Wed, Nov 11, 2009 at 9:24 AM, Roman Perepelitsa
> <[e-mail]> wrote:
> > 2009/11/11 Emanuele Rocci <[e-mail]>
> > > Hi All
> > > I am getting familiar with boost::lambda.
> > > Today I learnt about delayed variable and usage of boost::constant in
> > > lambda expression.
> > > Do you know why when I try to declare delayed variables for string like I
> > > do below
> > > the compiler gives me each time an error?
> > > I tried the following
> > > constant_type<const char*>::type _msg_a(constant("my msg"));
> > > constant_type<const std::string>::type _msg_b(constant("my msg"));
> > > constant_type<std::string>::type _msg_c(constant("my msg"));
> > > If I use constant( "my msg" ) inside a boost lambda expression everything
> > > is fine but I cannot declare correctly a delayed variable.
> > > Do you know how to fix it or where I can get some details?
> > > Thanks in advance
> > > Mn
> > >     
> > Function objects generated by lambda have unspecified types. If you want to
> > save them, use Boost.Function.
> > #include <iostream>
> > #include <boost/lambda/core.hpp>
> > #include <boost/function.hpp>
> > int main() {
> >   boost::function<const char*()> f = boost::lambda::constant("hello");
> >   std::cout << f() << std::endl;
> > }
> >   
>
> Or BOOST_AUTO (which will have lower overhead then Boost.Function).
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Could you translate the example to use BOOST_AUTO in place of
Boost.Function?

Ryan