Re: [Boost-users] [lambda] Using a reference in an expression

Subject: Re: [Boost-users] [lambda] Using a reference in an expression
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2012-09-13 00:07:28

AMDG

On 09/12/2012 12:14 PM, Bill Buklis wrote:
> How does one use a reference in a lambda expression? The documentation
> says that "var" does, but it clearly doesn't.
> 

var does work.  The argument to operator<<
is passed by reference correctly.  The
error is that the return type is deduced as
sample instead of sample&.  Try wrapping
the expression in bll::ret or look up
how to specify the return type in the
docs.  I don't remember off the top of my
head what you need to specialize.

> For example this fails to compile because it can't access the copy
> constructor:
> 
> #include <boost/lambda.hpp>
> namespace bll = boost::lambda;
> 
> class sample
> {
> public:
>     sample()
>     {
>     }
>     sample& operator<<( int )
>     {
>         return(*this);
>     }
> 
> private:
>     sample( const sample& );
> };
> 
> void test()
> {
>     typedef std::vector<int>    V;
> 
>     V        v;
>     sample    s;
> 
>     std::for_each(v.begin(), v.end(), bll::var(s) << bll::_1);
> };
> 

In Christ,
Steven Watanabe