Re: [Boost-users] Making an ordinary function into something useful in lambda expressions

From: Peter Waller (peter.waller_at_[hidden])
Date: 2007-10-12 15:30:32

Close, but not quite what I am after.

I would like to be able to define a function, say, cos, and have it
accept a functor, be it a var_type, and to return another one.

So I could write expressions like result = ( cos( _1 + 2 ) + 2 )(
value ), for example.

It would be nice if I could not have to write bind() where-ever I want
do this, otherwise the expressions will rapidly become unreadable.

TIA

- Peter

On 12/10/2007, Stuart Dootson <stuart.dootson_at_[hidden]> wrote:
> On 12/10/2007, Peter Waller <peter.waller_at_[hidden]> wrote:
> > I wish to write some lambda functions to have this sort of behaviour.
> >
> > #include <lambda/lambda.hpp>
> > #include <lambda/bind.hpp>
> > #include <function.hpp>
> >
> > using namespace boost::lambda;
> > using namespace boost;
> >
> > #include <iostream>
> > using std::cout; using std::endl;
> >
> > #include <cmath>
> >
> > int main( int argc, char *argv[] )
> > {
> >     float value;
> >     var_type<float>::type theValue( var( value ) );
> >
> >     // Ignore this line
> >     #define cos(theta) bind( unlambda(bind( cos, _1 )), theta )
> >
> >     function<float ( float )> myfunction = ( cos( theValue ) - 1 );
> >     value = 2;
> >     cout << "The result = " << myfunction(1) << endl;
> >     return 0;
> > }
> >
> > Namely, the cos( value ) part.
> >
> > The code above works, but I feel that using a #define is a little
> > messy, and seems hard to generalise to functions with more arguments.
> > I was wondering if there was a general, better way of turning a given
> > function into one whose execution is delayed.
> >
> > Thanks in advance,
> >
> > - Peter
>
> This seems to work:
>
>    double value = 3.14159 / 2.0;
>    boost::function<float(void)> fn =
> bind(static_cast<double(*)(double)>(&cos), var(value));
>    std::cout << fn() << std::endl;
>    value = 0.0;
>    std::cout << fn() << std::endl;
>
> At least, the output of 'fn()' changes when 'value' changes, which (as
> far as I could tell) was what you wanted?
>
> The static_cast is needed to select the appropriate 'cos' overload.
>
> HTH
>
> Stuart Dootson
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