Re: [Boost-users] return type of boost::bind

From: Cheenu Srinivasan (cheenu_srinivasan_at_[hidden])
Date: 2004-02-01 16:42:59

Peter Dimov wrote:
> Cheenu Srinivasan wrote:
> > Duane Murphy wrote:
> >> The return type of boost::bind is fairly indeterminate. In order to
> >> call a bound function at a later time, look at boost::function. It
> >> does a nice job of holding on to a function that uses boost::bind.
> > 
> > I looked at "Using bind with Boost.Function" in the boost::function
> > docs. Mimicing that, for my example, I tried:
> > 
> >   boost::function<double (int)> bound_func = boost::bind(f, _1, 1.234);
> >   int i = 99;
> >   cout << bound_func(i) << endl;
> > 
> > but it throws a bad_function_call exception when bound_func(i) is
> > invoked. How do I use boost::function for this situation?
> 
> Works for me; it's probably a compiler bug. Which compiler are you using?
> 
> Try
> 
> boost::function<double (int)> bound_func( boost::bind(f, _1, 1.234) );
> 
> instead.

With MSVC 7.1. the exception is thrown with both 'bound_func = bind(...)'
and 'bound_func(bind(...))'.  But I tried it with g++/solaris and it works.
So it does look like a bug with the MS compiler.  Thanks.