HI Roman!
Thanks for your nice response!
I have just discovered another way.
Please give me your comments!

constant_type< const char[7] >::type _name(constant("value: "));

playing the same role of
boost::function<const char*()> _name = boost::lambda::constant("value: ");
I find both 2 nice ways of expressing constant string in a lamda expression.

Using the boost::function approach, some sample code can be written like the following
using round brackets next to _name variable.
std::for_each( myList.begin(),myList.end(),(
        std::cout << _name() << _1  ,
         _1 = 2
            ));
I find the approach using boost::function very nice since it provides
more the "genericity" feeling; you don't have to declare the size of the
string but it is enough const char*().
With the constant_type way it is possible to use _name as a common const variable without
using round brackets
std::for_each( myList.begin(),myList.end(),(
        std::cout << _name << _1  ,
         _1 = 2
            ));
This works fine provided that the size is correct; it is probably not always nice, 
but compiler will complain if the size is not correct i.e. if you use const char[6]
or if you use const char[8].
Now I am wondering which is the fastest approach. I guess that probably they are
the same.
You commnnts or opinion are appreciated and off course thanks!
Manu

Date: Wed, 11 Nov 2009 17:24:35 +0100
From: roman.perepelitsa@gmail.com
To: boost-users@lists.boost.org
Subject: Re: [Boost-users] naming delayed variables string in boost lamdba

2009/11/11 Emanuele Rocci <rocciemanuele@hotmail.com>
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;
}

Roman Perepelitsa


Il tormentone dell'estate? Riascoltalo sulla Messenger Radio