2009-07-06
by earl, 5592 days ago
Sometimes, being accumulate(I->begin(), I->end(), 0, compose2(plus<int>(),
identity<int>(),
mem_fun(&C::F)));
Update: the above is nonsense (as in: won't compile). Problem is,
compose2 composes one binary and two unary functions such that:compose2(f, g, h)(x) = f(g(x), h(x)).What I want/need is a similar functional
F such that:F(f, g, h)(x, y) = f(g(x), h(y)).Any hints? How'd you call that, does it exist already? I know about Boost's rather versatile
bind, but that doesn't meet the point-free criterion.Update: In lack of a better name, let's call it
compose3:#include <functional>
template <typename F, typename G, typename H>
struct compose3_t : public std::binary_function<typename G::argument_type,
typename H::argument_type,
typename F::result_type> {
F f;
G g;
H h;
compose3_t(const F &f, const G &g, const H &h): f(f), g(g), h(h) {}
typename F::result_type operator()(const typename G::argument_type &x,
const typename H::argument_type &y) const {
return f(g(x), h(y));
}
};
template <typename F, typename G, typename H>
compose3_t<F, G, H> compose3(const F &f, const G &g, const H &h) {
return compose3_t<F, G, H>(f, g, h);
}
Now if we replace
compose2 with compose3 in the original example, it will compile and work as expected/desired.For illustration purposes, the same thing in Haskell:
compose3 :: (a -> b -> c) -> (u -> a) -> (v -> b) -> u -> v -> c
compose3 f g h x y = f (g x) (h y)
foldl (compose3 (+) id (f)) 0 v
Finally, let's let go of the insane hiding of points, but simply use lambdas (using boost::bind for the C++ version):
C++:accumulate(V->begin(), V->end(), 0, bind(plus<int>(), _1, bind(F, _2)))
Haskell:foldl (r x -> r + f x) 0 v
K:0 {x+f y}/v
If we relax things further, to allow a temporary vector being created, this simply collapses to:
Haskell:sum map (f) v
K:+/f v
Q:sum f v
Which concisely describes what the original piece of code was intended to achieve: sum a list element-wise transformed by a function.
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