Re: [Boost-users] [Bind] Understanding protect

Subject: Re: [Boost-users] [Bind] Understanding protect
From: Robert Jones (robertgbjones_at_[hidden])
Date: 2012-03-27 14:13:18

On Tue, Mar 27, 2012 at 3:46 PM, Igor R <boost.lists_at_[hidden]> wrote:

> > I thought its purpose was allow binding, but not evaluating, of a
> function object.
>
> ...when passed to bind() as a not first argument.
>
> Ah, right, thx.

>
> > So that this should work.
> >
> > #include <iostream>
> > #include <boost/bind.hpp>
> > #include <boost/bind/protect.hpp>
> >
> > void f( int i ) { std::cout <<  i << std::endl; }
> >
> > template <typename NullaryCallable> void g( const NullaryCallable &
> callable
> > ) { callable( ); }
> >
> > int main( )
> > {
> >     g( protect( bind( f, _1 ) )( 1 ) );
> > }
> >
> > But it doesn't, and instead thinks I'm calling g() with the return type
> of
> > f(), just as if I'd not used protect at all.
>
> Sure, protect doesn't do anything here.
>
> > Is it possible to have bind generate a fully bound, nullary callable
> object?
>
> bind(f, 1);
>

 Yeah, but that doesn't quite fit the bill! Ultimately I'm doing this for
each member of a
range, via std algorithm, so I need to bind a unary argument to produce a
nullary function
object, which I can then pass to an evaluation context, ie my g( )
function. If I've generated
the function object with bind and placeholder(s), it seems there's no way
to prevent the binding
of the last placeholder from triggering evaluation, when what I require is
an unevaluated nullary
function object.

Thx, Rob.