1

我很难使用具有以下问题的可变参数模板。

假设所有谓词函子的形式为，

class Pred1 {
public:
    Pred1( Args... ); // The signature Args... can vary class to class.

    template <typename T>
    bool operator()(T t);
};

鉴于这些函子，我想制作一个可变参数模板类，如果每个谓词的所有 operator() 都返回 true，则返回 true，即，

template <typename... Preds>
class CombinePredAnd {
public:
    template <typename T>
    bool operator()(T t){
         // returns true if all of the Preds( Args... ).operator()(t) returns true;  
         // Args... should be passed when CombinePredAnd is constructed.
    }
};

对我来说，我不知道将参数传递给 Preds 的每个构造函数。你能给我一些提示吗？另外，如果你有更好的设计和相同的功能，请告诉我。

4

2 回答 2

5

也许是这样的：

#include <tuple>
#include <type_traits>

template <std::size_t N, std::size_t I, typename Tuple>
struct evaluate_all
{
    template <typename T>
    static bool eval(T const & t, Tuple const & preds)
    {
        return std::get<I>(preds)(t)
            && evaluate_all<N, I + 1, Tuple>::eval(t, preds);
    }
};

template <std::size_t N, typename Tuple>
struct evaluate_all<N, N, Tuple>
{
    template <typename T>
    static bool eval(T const &, Tuple const &)
    {
        return true;
    }
};

template <typename ...Preds>
struct conjunction
{
private:
    typedef std::tuple<Preds...> tuple_type;
    tuple_type preds;

public:
    conjunction(Preds const &... p) : preds(p...) { }

    template <typename T>
    bool operator()(T const & t) const
    {
        return evaluate_all<sizeof...(Preds), 0, tuple_type>::eval(t, preds);
    }
};

template <typename ...Preds>
conjunction<typename std::decay<Preds>::type...> make_conjunction(Preds &&... preds)
{
    return conjunction<typename std::decay<Preds>::type...>(std::forward<Preds>(preds)...);
}

用法：

auto c = make_conjunction(MyPred(), YourPred(arg1, arg2, arg3));

if (c(10)) { /* ... */ }

例子：

#include <iostream>

typedef int T;

struct Pred1
{
    int a;
    Pred1(int n) : a(n) { }
    bool operator()(int n) const { return n >= a; }
};
struct Pred2
{
    int a;
    Pred2(int n) : a(n) { }
    bool operator()(int n) const { return n <= a; }
};

int main()
{
    auto c = make_conjunction(Pred1(1), Pred2(3));

    std::cout << "1: " << c(1) << "\n"
              << "5: " << c(4) << "\n";
}

注意：您也可以使此方法的“合取”部分参数化，因此您只需插入std::logical_andor即可拥有合取和析取std::logical_or。

于 2012-11-25T12:22:00.913 回答
0

这是 Kerrek SB 的替代解决方案：

#include <tuple>
#include <type_traits>
#include <cstdlib>

template <typename HeadPred, typename ...TailPreds>
struct CombinePredAnd 
{
    template<typename H, typename ...Ts>
    explicit CombinePredAnd(H const & h, Ts const &...ts)
    : _preds(h, ts...){}

    template <typename T>
    bool operator()(T t){
        return eval(t,_preds);
    }

private:

    template<typename T, size_t I = 0, typename ...Ps>
    typename std::enable_if<sizeof ...(Ps) == I,bool>::type
    static eval(T t, std::tuple<Ps...>) {
        return true;
    }

    template<typename T, size_t I = 0, typename ...Ps>
    typename std::enable_if<sizeof ...(Ps) != I,bool>::type
    static eval(T t, std::tuple<Ps...> const & preds) {
        auto const & pred = std::get<I>(preds);
        return pred(t) && eval<T,I + 1>(t,preds);
    }

    std::tuple<HeadPred, TailPreds...> _preds;
};

这可以按照 Kerrek SB 建议的任意谓词函子的通用合取或析取的方式概括如下，通过选择合取或析取来参数化。附加了一个测试程序，使用 gcc 4.7.2 和 clang 3.2 构建：

#include <tuple>
#include <type_traits>
#include <functional>
#include <cstdlib>

template <class AndOrOr, typename HeadPred, typename ...TailPreds>
struct dis_or_con_join 
{
    static_assert(
        std::is_same<AndOrOr,std::logical_and<bool>>::value ||
        std::is_same<AndOrOr,std::logical_or<bool>>::value,     
        "AndOrOr must be std::logical_and<bool> or std::logical_or<bool>");

    template<typename H, typename ...Ts>
    explicit dis_or_con_join(H const & h, Ts const &...ts)
    : _preds(h, ts...){}

    template <typename T>
    bool operator()(T t){
        return eval(t,_preds);
    }

private:

    static const bool conjunction = 
        std::is_same<AndOrOr,std::logical_and<bool>>::value;

    template<typename T, size_t I = 0, typename ...Ps>
    typename std::enable_if<sizeof ...(Ps) == I,bool>::type
    static eval(T t, std::tuple<Ps...>) {
        return conjunction;
    }

    template<typename T, size_t I = 0, typename ...Ps>
    typename std::enable_if<sizeof ...(Ps) != I,bool>::type
    static eval(T t, std::tuple<Ps...> const & preds) {
        auto lamb = conjunction ? 
            [](bool b){ return b; } :
            [](bool b){ return !b; };               
        auto const & pred = std::get<I>(preds);
        return lamb(lamb(pred(t)) && lamb(eval<T,I + 1>(t,preds)));
    }

    std::tuple<HeadPred, TailPreds...> _preds;
};

template<typename HeadPred, typename ...TailPreds>
using conjunction  = 
dis_or_con_join<std::logical_and<bool>,HeadPred,TailPreds...>;

template<typename HeadPred, typename ...TailPreds>
using disjunction =
dis_or_con_join<std::logical_or<bool>,HeadPred,TailPreds...>;


// Test...

#include <iostream>
#include <list>
#include <algorithm>

using namespace std;

// For various predicates with various constructors...

template<typename T>
struct is_in_list
// Stores an arbitrary sized list of its type T constructor arguments
// and then with tell us whether any given T is in its list 
{
    is_in_list(initializer_list<T> il)
    : _vals(il.begin(),il.end()){}
    bool operator()(T t) const {
        return find(_vals.begin(),_vals.end(),t) != _vals.end();
    }
    list<T> _vals;
};

int main()
{
    is_in_list<char> inl03 = {'\0','\3'};
    is_in_list<long> inl013 = {0,1,3};
    is_in_list<float> inl0123 = {0.0f,1.0f,2.0f,3.0f};

    conjunction<is_in_list<char>,is_in_list<long>,is_in_list<float>> 
    conj{inl03,inl013,inl0123};
    disjunction<is_in_list<char>,is_in_list<long>,is_in_list<float>> 
    disj{inl03,inl013,inl0123};

    cout << "conjunction..." << endl;
    cout << 1 << " is " << (conj(1) ? "" : "not ") 
        << "in all the lists" << endl;
    cout << 0 << " is " << (conj(0) ? "" : "not ") 
        << "in all the lists" << endl;
    cout << 3 << " is " << (conj(3) ? "" : "not ") 
        << "in all the lists" << endl;
    cout << "disjunction..." << endl;       
    cout << 1 << " is in " << (disj(1) ? "at least one " : "none ") 
        << "of the lists" << endl;
    cout << 2 << " is in " << (disj(2) ? "at least one " : "none ") 
        << "of the lists" << endl;
    cout << 3 << " is in " << (disj(3) ? "at least one " : "none ") 
        << "of the lists" << endl;
    cout << 4 << " is in " << (disj(4) ? "at least one " : "none ") 
        << "of the lists" << endl;
    return 0;
}

可能不清楚的线：

return lamb(lamb(pred(t)) && lamb(eval<T,I + 1>(t,preds)));

只是利用等价性：

P or Q = not(not(P) and not(Q))

输出：-

conjunction...
1 is not in all the lists
0 is in all the lists
3 is in all the lists
disjunction...
1 is in at least one of the lists
2 is in at least one of the lists
3 is in at least one of the lists
4 is in none of the lists
于 2013-04-08T10:11:17.990 回答