using of std::accumulate

Question

Need prettier solution of below example but with std::accumulate.

#include 
#include 
#include 

class Object
{
pu         


        

Answer 1

Update 2: Boost.Lambda makes this a piece of cake:

// headers
#include 
#include 
using namespace boost::lambda;
// ...
cout << accumulate(dv.begin(), dv.end(), 
                   0, 
                   _1 += bind(&strange::value, _2)) //strange defined below
     << endl;

Update: This has been bugging me for a while. I can't just get any of the STL algorithms to work in a decent manner. So, I rolled my own:

// include whatever ...
using namespace std;

// custom accumulator that computes a result of the 
// form: result += object.method();
// all other members same as that of std::accumulate
template 
V accumulate2(I first, I last, V val, Fn1 op, Fn2 memfn) {
    for (; first != last; ++first)
        val = op(val, memfn(*first));
    return val;
}

struct strange {
    strange(int a, int b) : _a(a), _b(b) {}
    int value() { return _a + 10 * _b; }
    int _a, _b;
};

int main() {
    std::vector dv;
    dv.push_back(strange(1, 3));
    dv.push_back(strange(4, 6));
    dv.push_back(strange(20, -11));        
    cout << accumulate2(dv.begin(), dv.end(), 
                        0, std::plus(), 
                        mem_fun_ref(&strange::value)) << endl;
}

Of course, the original solution still holds: The easiest is to implement an operator+. In this case:

double operator+(double v, Object const& x) {
        return v + x.a_;
}

and make it a friend of Object or member (look up why you may prefer one over the other):

class Object
{
   //...
  friend double operator+(double v, Object const& x);

and you're done with:

 result = accumulate(collection.begin(), collection.end(), 0.0);

My earlier approach doesn't work because we need a binary_function.

std::accumulate manual.