为什么bind1st和bind2nd需要常量函数对象？

Question

So, I was writing a C++ program which would allow me to take control of the entire world. I was all done writing the final translation unit, but I got an error:

error C3848: expression having type 'const `anonymous-namespace'::ElementAccumulator<T,BinaryFunction>' would lose some const-volatile qualifiers in order to call 'void `anonymous-namespace'::ElementAccumulator<T,BinaryFunction>::operator ()(const point::Point &,const int &)'
        with
        [
            T=SideCounter,
            BinaryFunction=std::plus<int>
        ]
        c:\program files (x86)\microsoft visual studio 9.0\vc\include\functional(324) : while compiling class template member function 'void std::binder2nd<_Fn2>::operator ()(point::Point &) const'
        with
        [
            _Fn2=`anonymous-namespace'::ElementAccumulator<SideCounter,std::plus<int>>
        ]
        c:\users\****\documents\visual studio 2008\projects\TAKE_OVER_THE_WORLD\grid_divider.cpp(361) : see reference to class template instantiation 'std::binder2nd<_Fn2>' being compiled
        with
        [
            _Fn2=`anonymous-namespace'::ElementAccumulator<SideCounter,std::plus<int>>
        ]

I looked in the specifications of binder2nd and there it was: it took a const AdaptibleBinaryFunction.

So, not a big deal, I thought. I just used boost::bind instead, right?

Wrong! Now my take-over-the-world program takes too long to compile ( bind is used inside a template which is instantiated quite a lot)! At this rate, my nemesis is going to take over the world first! I can't let that happen -- he uses Java!

So can someone tell me why this design decision was made? It seems like an odd decision. I guess I'll have to make some of the elements of my class mutable for now...

EDIT: The offending code:

template <typename T, typename BinaryFunction>
class ElementAccumulator 
    : public binary_function<typename T::key_type, typename T::mapped_type, void>
{
public:
    typedef T MapType;
    typedef typename T::key_type KeyType;
    typedef typename T::mapped_type MappedType;
    typedef BinaryFunction Func;

    ElementAccumulator(MapType& Map, Func f) : map_(Map), f_(f) {}

    void operator()(const KeyType& k, const MappedType& v)
    {
        MappedType& val = map_[k];
        val = f_(val, v);
    }
private:
    MapType& map_;
    Func f_;
};

void myFunc(int n)
{
    typedef boost::unordered_map<Point, int, Point::PointHash> Counter;
    Counter side_count;
    ElementAccumulator<SideCounter, plus<int> > acc(side_count, plus<int>());

        vector<Point> pts = getPts();
    for_each(pts.begin(), pts.end(), bind2nd(acc, n));
}

Answer 1

binder2nd ctor takes a constant reference to an AdaptableBinaryFunction -- not a const AdaptableBinaryFunction per se. How's your instantiating-code? One normally doesn't explicitly mention binder2nd but rather work through convenience function bind2nd (which simply works on the second argument x with a typename Operation::second_argument_type(x) or the like).

Answer 2

Well, attempting some deduction:

The reason anything takes a const anything, is to permit someone to pass a const anything into it.

The most obvious const something that you want to pass into "functional" functions is a reference to a temporary.

In particular, if bind1st and other stuff in <functional> took a non-const reference parameter, then you couldn't chain them together to program in a functional style. A functional style abhors the idea of capturing a temporary in a variable in one statement, and then "later" modifying that variable in "the next" statement. All very imperative and side-effecty.

Unfortunately, this means that as <functional> is defined, operator() of functors needs to be const in this case and presumably a bunch of other cases. Yours isn't.

Does boost::bind allow either const or not as part of the template type where relevant? If so then maybe <functional> doesn't do this simply because boost::bind was designed once people had more idea how to get the best out of templates. Or maybe bind1st was designed with a purer functional mindset, hence no side-effects, hence why shouldn't everything be const? I may have missed part of the point of the question - I see from your code example why you want to use parameter binding, but I don't think it's obvious that a header called <functional> is the right place to look for anything involving accumulators ;-)

Answer 3

bind (and the old, deprecated bind1st and bind2nd ) are value-semantic. The return object is self-contained and it doesn't reference the parameters, const or not.

To get reference semantics instead, pass std::ref to bind .

auto defer_by_value = std::bind( fun, foo, bar ); // Copy fun, foo, and bar.

auto defer_by_ref = std::bind( std::ref( fun ), std::ref( foo ), std::ref( bar ) );
                                                  // Observe fun, foo, and bar.