From: David Morrison (dave_at_[hidden])
Date: 2005-03-22 15:28:55
Pablo Aguilar wrote:
> Maybe not the best solution performance-wise, but you can try the following:
>
> Replace
> typedef float (A::*pfn)(int, float);
> with
> typedef boost::function3<float,A*,int,float> pfn;
Thanks, that works nicely, though the required coding difference is a
bit disconcerting. I found that if I created a class in analogy with
"apply" called, say, "apply_mem_fn" like this:
namespace boost
{
template<class R> struct apply_mem_fn
{
typedef R result_type;
template<class F, class A1, class A2, class A3> result_type
operator()(F & f, A1 & a1, A2 & a2, A3 & a3) const
{
return ((a1).*(f))(a2, a3);
}
};
} // namespace boost
Then I can achieve what I was trying to accomplish with code that looks
nearly the same as in the case of pointers to non-member functions:
#include <vector>
#include <iostream>
#include <iterator>
#include "boost/bind.hpp"
namespace boost
{
template<class R> struct apply_mem_fn
{
typedef R result_type;
template<class F, class A1, class A2, class A3> result_type
operator()(F & f, A1 & a1, A2 & a2, A3 & a3) const
{
return ((a1).*(f))(a2, a3);
}
};
} // namespace boost
using namespace std;
using namespace boost;
struct A {
float a(int m, float x) { return m * x;}
float b(int m, float x) { return m + x;}
float c(int m, float x) { return m - x;}
float d(int m, float x) { return m / x;}
};
typedef float (A::*pfn)(int, float);
int
main()
{
A a;
int m = 2;
float x = 3.0;
vector<pfn> v;
vector<float> f;
v.push_back(&A::a);
v.push_back(&A::b);
v.push_back(&A::c);
v.push_back(&A::d);
transform(v.begin(), v.end(), back_inserter(f),
bind(apply_mem_fn<float>(), _1, a, m, x));
copy(f.begin(), f.end(), ostream_iterator<float>(cout, "\n"));
}
It's not clear to me whether it's possible to choose the right
implementation based on the argument.
Dave