[Boost-users] Re: boost::function

30 Oct 2004


      sboschi75-personal@yahoo.it wrote:
...
Hallo,
I would like to override operators for boost::function.
Hi Stefano,
in future, please use more specific subject for your postings. Also, putting
your full name in the "From" line is considered more polite than using
"sboschi75"
...
Forexample I was thinking of an operator "-" :
template <T>
function<T> operator - (const function<T>& a) {
         // what do I write here in roder to return the symmetric
         function?
}
The following works for me.


#include 
#include 
#include <functional>
#include <iostream>

using namespace std;

template<class T>
boost::function<T> operator-(boost::function<T> a)
{
    typedef typename boost::function<T>::result_type rt;
    return boost::bind(std::minus<rt>(), rt(),
                boost::bind(a, _1));
}

int foo(int i)
{
    return i + 2;
}

int main()
{
    boost::function f = foo;
    boost::function f2 = -f;
    cout << f(4) << " " << f2(4) << "\n";
    return 0;
}

- Volodya

[Boost-users] Re: boost::function

Vladimir Prus