Re: [Boost-users] [mem_fn] return type

1 Sep 2009

      On Tue, Sep 1, 2009 at 1:57 AM, OvermindDL1  wrote:
...
On Mon, Aug 31, 2009 at 9:33 PM, Lowell McLinskey wrote:
...
I am having trouble understanding the return type of mem_fn.  Is it possible
to create a vector of this type, analogous to how one can create a vector of
type mem_fun_t to hold the return values of mem_fun()?  Here is an example
of what I'd like to do.  Thanks in advance.
#include <vector>
#include <functional>
#include "boost/shared_ptr.hpp"
#include "boost/mem_fn.hpp"
class A
{
public:
    A(const int a) : a_(a) {}
    const int get() const {return a_;}
private:
    int a_;
};
int main(int argc, char** argv)
{
    // I can do this
    std::vector > v;
    v.push_back(std::mem_fun(&A::get));
    int i = v[0](new A(4));
    // and this
    boost::shared_ptr<A> p(new A(7));
    int j = boost::mem_fn(&A::get)(p);
    // but how do I do this?
    // std::vector< WHAT GOES HERE > w;
    // w.push_back(boost::mem_fn(&A::get));
    // int k = w[0](p);
    return 0;
}
I do not know if mem_fun supports the boost::result_of template, but
if it does that would let you get what you need, and if it does not
support it then that needs to be fixed.
It looks like the documentation indicates that boost::result_of is supported:

unspecified-N::result_type is defined as the return type of the member
function pointer passed as an argument to mem_fn (R in the following):

template unspecified-2 mem_fn(R (T::*pmf) () const);

I'm still not sure what the template parameter should be to make the
following work:

    typedef boost::function_types::result_type< What Goes Here > result_type;
    std::vector w;
    w.push_back(boost::mem_fn(&A::get));
    boost::shared_ptr<A> p(new A(7));
    int k = w[0](p); // k = 7

I was able to deduce from compiler errors something that does actually
work, but it doesn't seem very intuitive:

    boost::shared_ptr<A> p(new A(7));
    std::vector > u;
    u.push_back(boost::mem_fn(&A::get));
    int n = u[0](p); // n = 7