On Wed, Nov 11, 2009 at 9:24 AM, Roman Perepelitsa wrote:

2009/11/11 Emanuele Rocci

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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::type _msg_a(constant("my msg")); constant_type<const std::string>::type _msg_b(constant("my msg")); constant_typestd::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 #include int main() {   boost::function f = boost::lambda::constant("hello");   std::cout << f() << std::endl; }

Or BOOST_AUTO (which will have lower overhead then Boost.Function).

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 ofboost::function _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 bracketsstd::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!ManuDate: 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 Hi AllI 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::type _msg_a(constant("my msg"));constant_type<const std::string>::type _msg_b(constant("my msg")); constant_typestd::string::type _msg_c(constant("my msg")); If I use constant( "my msg" ) inside a boost lambda expression everythingis 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 advanceMn Function objects generated by lambda have unspecified types. If you want to save them, use Boost.Function. #include <iostream>#include #include int main() { boost::function f = boost::lambda::constant("hello"); std::cout << f() << std::endl;} Roman Perepelitsa _________________________________________________________________ Scrivi, parla, gioca... Scopri Messenger 2009 http://www.messenger.it/home_comunica.aspx