Re: [Boost-users] break and return statement in lambda expression

Subject: Re: [Boost-users] break and return statement in lambda expression
From: Conoscenza Silente (abruzzoforteegentile_at_[hidden])
Date: 2010-07-22 08:06:14

Well
I was willing to do 2 things
*  to use call 'break' to get out of the std::for_each
*  to use return to exit out of the class function having this code.
For the 'break' command now I found other means ( shame on me: I can't use
break in a std::for_each like if ..typed without too much thinking ..sorry
)  but still I don't know how to fix the call to 'return'.
If I put a call to 'return' in lambda expression inside std::for_each to
exit the function I have the same error.

On Thu, Jul 22, 2010 at 1:45 PM, OvermindDL1 <overminddl1_at_[hidden]> wrote:

> On Thu, Jul 22, 2010 at 5:36 AM, Conoscenza Silente
> <abruzzoforteegentile_at_[hidden]> wrote:
> > Hi All
> > I created a lambda expression inside my std::for_each call.
> > Inside it there is code like this one, but I have building error telling
> me
> > that
> >
> > error: expected primary-expression before ‘break’
> > error: expected `]' before ‘break’
> >
> > Do you know what it is and how to fix it?
> > The same applies to return statement
> > Thanks
> > AFG
> >
> > namespace bl = boost::lambda;
> > int a, b;
> > bl::var_type< int >::type a_( bl::var( a ) );
> > bl::var_type< int >::type b_( bl::var( b ) );
> >
> > std::for_each( v.begin(), v.end(), (
> > if_(  a_ > _b_ )
> > [
> > std::cout << _1,
> >  break
> > ]
> > ));
>
> 'break' is not a lazy command, hence it will not work there.  What are
> you even trying to do anyway?
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