Question

是否有办法部分绑定可调用对象（例如函数）的第一个/最后一个 n 参数而不显式指定其余参数？

std::bind()似乎要求所有参数被绑定，那些要留下的参数应绑定到std::placeholders::_1，_2，{{ 1}}等。

是否可以从第一个/最后一个参数开始为部分绑定编写_3 / bind_first()，并且可以在原始位置以原始顺序自动为任何剩余的未绑定参数插入占位符？

Answer 1

Boost和标准库bind都不会自动填充空白。如果你有一个下雨的夜晚，你可以自己写这样的小工具;这里只是一个简单函数的尾随参数的示例：

#include <tuple>
#include <type_traits>
#include <utility>

template <typename F, typename ...Args> struct trailing_binder;

template <typename R, typename ...Frgs, typename ...Args>
struct trailing_binder<R(Frgs...), Args...>
{
    template <typename ...Brgs>
    trailing_binder(R (*f)(Frgs...), Brgs &&... brgs)
    : the_function(f)
    , the_args(std::forward<Brgs>(brgs)...)
    { }

    template <unsigned int ...I> struct intlist {};

    template <typename ...Brgs>
    typename std::enable_if<sizeof...(Brgs) + sizeof...(Args) == sizeof...(Frgs), R>::type
    operator()(Brgs &&... brgs)
    {
        return unwrap(std::integral_constant<bool, 0 == sizeof...(Args)>(),
                      intlist<>(),
                      std::forward<Brgs>(brgs)...);
    }

private:
    template <unsigned int ...I, typename ...Brgs>
    R unwrap(std::false_type, intlist<I...>, Brgs &&... brgs)
    {
        return unwrap(std::integral_constant<bool, sizeof...(I) + 1 == sizeof...(Args)>(),
                      intlist<I..., sizeof...(I)>(),
                      std::forward<Brgs>(brgs)...);
    }

    template <unsigned int ...I, typename ...Brgs>
    R unwrap(std::true_type, intlist<I...>, Brgs &&... brgs)
    {
        return the_function(std::get<I>(the_args)..., std::forward<Brgs>(brgs)...);
    }

    R (*the_function)(Frgs...);
    std::tuple<Args...> the_args;
};

template <typename R, typename ...Args, typename ...Frgs>
trailing_binder<R(Frgs...), Args...> trailing_bind(R (*f)(Frgs...), Args &&... args)
{
    return trailing_binder<R(Frgs...), typename std::decay<Args>::type...>(f, std::forward<Args>(args)...);
}

用法：

int f(int a, int b, int c, int d) { return a + b + c + d; }

int main()
{
    auto b = trailing_bind(f, 1);
    return b(3, 8, 13);
}

Answer 2

受到这个问题的启发，我从头开始编写自己的预绑定。它最终看起来与其他人非常相似，但我保证它是原创的:) - 称之为融合进化。

虽然味道略有不同。首先，它转发给它的构造函数，但你可能更喜欢使用std::decay（它在某些方面更有意义，但我不喜欢在任何地方写std::ref。另一方面，我添加了对嵌套预绑定的支持，因此prebind(foo, prebind(GetRandomNumber))()与prebind(foo)(GetRandomNumber())相同。

#include <tuple>
#include <type_traits>
using namespace std;

struct pb_tag {}; //use inheritance to mark prebinder structs

//result_of_t will be defined by default in c++1y
template<typename T > using result_of_t = typename result_of<T>::type;
template<typename T> using is_prebinder = is_base_of<pb_tag, typename remove_reference<T>::type >;

//ugly sequence generators for something different
template<int N, int ...S> struct seq : seq<N-1, N, S...> {};
template<int ...S> struct seq<0, S...> {typedef seq type;};

//these three functions are only for nested prebind. they map
//T t -> T t and Prebind<f, T...> -> f(T...)
template<typename T>
auto dispatchee(T&& t, false_type) -> decltype(forward<T>(t)){
    return forward<T>(t);
}

template<typename T>
auto dispatchee(T&& t, true_type) -> decltype(t())
{
    return t();
}

template<typename T>
auto expand(T&& t) -> decltype(dispatchee(forward<T>(t), is_prebinder<T>()))
{
    return dispatchee(forward<T>(t), is_prebinder<T>());
}

template<typename T> using expand_type = decltype(expand(declval<T>()));

//the functor which holds the closure in a tuple
template<typename f, typename ...ltypes>
struct prebinder : public pb_tag
{
    tuple<f, ltypes...> closure;
    typedef typename seq<sizeof...(ltypes)>::type sequence;
    prebinder(f F, ltypes... largs) : closure(F, largs...) {}

    template<int ...S, typename ...rtypes>
    result_of_t<f(expand_type<ltypes>..., rtypes...)>
    apply(seq<0, S...>, rtypes&& ... rargs){
        return (get<0>(closure))(expand(get<S>(closure))... , forward<rtypes>(rargs)...);
    }

    template<typename ...rtypes>
    result_of_t<f(expand_type<ltypes>..., rtypes...)>
    operator() (rtypes&& ... rargs){
        return apply(sequence(), forward<rtypes>(rargs)...);
    }
};

template<typename f, typename ...ltypes>
prebinder<f, ltypes...> prebind(f&& F, ltypes&&... largs)
{
    return prebinder<f, ltypes...>(forward<f>(F), forward<ltypes>(largs)...);
}

它也可以轻松更改为postbind。

用法如下：

int g(int a){ return 1 + a; }

int h(){ return 1; }

int i(int a, int b, int c, int d){
    return 1 + a + b + c + d;
}

int main()
{
    //completely bound
    auto a = prebind(g, 1);
    cout << a() << endl;

    //nested bind by reference
    auto b = prebind(g, a);
    cout << b() << endl;
    get<1>(a.closure) = 2;
    cout << b() << endl;

    //bind to prebinder
    auto c = prebind(b);
    cout << c() << endl;

    //nested bind of temp to temp
    auto d = prebind(prebind(g), prebind(h));
    cout << d() << endl;

    //and the one you wanted orginally
    auto e = prebind(i, 1, 1, 1);
    cout << e(1) << endl;

    return 0;
}

Answer 3

现在的建议是避免Bind，并使用lambda包装对成员函数的调用。

https://releases.llvm.org/10.0.0/tools/clang/tools/extra/docs/clang-tidy/checks/modernize-avoid-bind.html

函数参数的部分绑定

3 个答案: