naming delayed variables string in boost lamdba

[manustone]

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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<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

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[Roman Perepelitsa-3]

2009/11/11 Emanuele Rocci <rocciemanuele@...>

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

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[OvermindDL1]

On Wed, Nov 11, 2009 at 9:24 AM, Roman Perepelitsa
<roman.perepelitsa@...> wrote:

> 2009/11/11 Emanuele Rocci <rocciemanuele@...>
>>
>> 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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[Ryan McConnehey]

OvermindDL1 wrote:

On Wed, Nov 11, 2009 at 9:24 AM, Roman Perepelitsa
roman.perepelitsa@... wrote:
  

2009/11/11 Emanuele Rocci rocciemanuele@...
    

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

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[manustone]

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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@...
To: boost-users@...
Subject: Re: [Boost-users] naming delayed variables string in boost lamdba

2009/11/11 Emanuele Rocci <rocciemanuele@...>

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

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[Roman Perepelitsa-3]

2009/11/11 Emanuele Rocci <rocciemanuele@...>

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
            ));

This does not do what you think it does. It'll evaluate and output _name() only once. Here's how you can fix it:

std::for_each( myList.begin(),myList.end(),(

        std::cout << bind(_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 as expected -- _name is being output several times.

 

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. 

Using constant_type should be faster, because it does not use virtual functions. Also, to my surprise, it's a documented way of creating lazy constants (see http://www.boost.org/doc/libs/1_40_0/doc/html/lambda/le_in_details.html#lambda.delaying_constants_and_variables).

Roman Perepelitsa.

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