Re: [Boost-users] How can I save the values in lambda expression? Can any one with kindness help me?

As you know, I am a Chinese and according to my knowledge few book in Chinese introduce boost library. I only have the book " Beyond the C++ Standard Library: An introduction to boost" (written by Bjoern Karlsson, translated in Chinese).Only several pages said about lambda. I tried to ask www.csdn.net( a Chinese developers' net) and few people answered my questions. I have not studied the source code of lambda yet. So all I have to do is to ask www.boost.org to get answers and go forth. I think the var(switchvalue) is a pointer to a function from the example code in the book: .......... std::for_each(vec.begin(),vec.end(),var(m)=_1); .............. So the operations in m can be performed. Am I right? But I do not know how to save the values in lambda expression. Thanks a lot. 在2009-07-09，"Steven Watanabe" 写道：

AMDG

fmingu wrote:

...

I am using Dev-C++ with boost lambda expression, I wrote: .................... int nnumber; int switchvalue; typedef std::map IIPrimemap; IIPrimemap primemapvec; # define BetaKGre2 1 # define EqualToOne 2 #define ProductOfKPrime 3 ..................... if_then_else(var(nnumber)==1,var(switchvalue)=EqualToOne, if_then_else(constant(count_if(primemapvec.begin(),primemapvec.end(), bind(&IIPrimemap::value_type::second,_1)>2)>0), var(switchvalue)=BetaKGre2,var(switchvalue)=ProductOfKPrime)); std::cout<<" the var(switchvalue) is "<

without any expression of var(switchvalue). I do not know how to save the values in lambda expression and where is wrong? Can I use *var(switchvalue) instead of var(switchvalue)? Can anyone with kindness help me?

Think about it. What exactly is var(switchvalue)? What does std::cout << var(switchvalue) do?

In Christ, Steven Watanabe