C++元程序自动创建函数

C++ metaprogramming automatic function creation?

本文关键字：创建 函数 程序 C++      更新时间：2023-10-16

我不确定标题是否正确，但这是我的问题：

我想使用元编程来为特定的表达式创建函数。例如，假设我们有这样的代码：

template<typename T1, typename T2>
struct plus{
    T1 func(T1 in1, T2 in2){ return in1 + in2; }
};
template<typename T1, typename T2, typename T3, typename expr>
struct wrap{
    /* contain a func that can evaluate the expr */
};

程序员将编写下面的代码，以便为表达式创建一个函数：

wrap<int,int,int,plus<plus<int,int>,int> >::func(1,2,3); /*result should be 6*/

这可能吗？

谢谢。

#include <utility>
#include <tuple>
#include <cstddef>
struct arg
{
    template <typename Arg1>
    static constexpr decltype(auto) apply(Arg1&& arg1)
    {
        return std::forward<Arg1>(arg1);
    }
    static constexpr std::size_t arity = 1;
};
template <typename Type, Type value>
struct constant
{    
    static constexpr decltype(auto) apply()
    {
        return value;
    }
    static constexpr std::size_t arity = 0;
};
template <typename Lhs, typename Rhs>
struct plus
{
    template <typename... Args>
    static constexpr decltype(auto) apply(Args&&... args)
    {
        return _apply(std::make_index_sequence<Lhs::arity>{}, std::make_index_sequence<Rhs::arity>{}, std::tuple<Args&&...>(std::forward<Args>(args)...));
    }
    template <typename Tuple, std::size_t... Arity1, std::size_t... Arity2>
    static constexpr decltype(auto) _apply(std::index_sequence<Arity1...>, std::index_sequence<Arity2...>, Tuple&& args)
    {
        return Lhs::apply(static_cast<typename std::tuple_element<Arity1, Tuple>::type>(std::get<Arity1>(args))...)
             + Rhs::apply(static_cast<typename std::tuple_element<Lhs::arity + Arity2, Tuple>::type>(std::get<Lhs::arity + Arity2>(args))...);
    }
    static constexpr std::size_t arity = Lhs::arity + Rhs::arity;
};
template <typename Lhs, typename Rhs>
struct multiply
{
    template <typename... Args>
    static constexpr decltype(auto) apply(Args&&... args)
    {
        return _apply(std::make_index_sequence<Lhs::arity>{}, std::make_index_sequence<Rhs::arity>{}, std::tuple<Args&&...>(std::forward<Args>(args)...));
    }
    template <typename Tuple, std::size_t... Arity1, std::size_t... Arity2>
    static constexpr decltype(auto) _apply(std::index_sequence<Arity1...>, std::index_sequence<Arity2...>, Tuple&& args)
    {
        return Lhs::apply(static_cast<typename std::tuple_element<Arity1, Tuple>::type>(std::get<Arity1>(args))...)
             * Rhs::apply(static_cast<typename std::tuple_element<Lhs::arity + Arity2, Tuple>::type>(std::get<Lhs::arity + Arity2>(args))...);
    }
    static constexpr std::size_t arity = Lhs::arity + Rhs::arity;
};

测试：

int main()
{
    // (1 + 2) + 3 = 6
    std::cout << plus<plus<arg, arg>, arg>::apply(1, 2, 3) << std::endl;
    // (a + 5) + (2 * 6) = 9 + 12 = 21
    int a = 4;
    std::cout << plus<plus<arg, arg>, multiply<arg, constant<int, 6>>>::apply(a, 5, 2) << std::endl;
    // ((1 * 2) * 3) * 4 = 24
    std::cout << multiply<multiply<multiply<arg, arg>, arg>, arg>::apply(1, 2, 3, 4) << std::endl;
    // 2 + (4 * 5) = 22
    static_assert(plus<arg, multiply<arg, arg>>::apply(2, 4, 5) == 22, "!");
}

输出：

6
21
24

演示1

上面的解决方案可以改进，这样引入新的函子需要更少的精力，声明本身也更可读，如下所示：

#include <iostream>
#include <utility>
#include <tuple>
#include <cstddef>
template <std::size_t Arity>
struct expression
{    
    static constexpr std::size_t arity = Arity;
};
template <typename Expr, typename Rhs>
struct unary_expression : expression<Rhs::arity>
{    
    template <typename... Args>
    static constexpr decltype(auto) apply(Args&&... args)
    {
        static_assert(sizeof...(Args) == unary_expression::arity, "Wrong number of operands!");
        return Expr::eval(Rhs::apply(std::forward<Args>(args)...));
    }
};
template <typename Expr, typename Lhs, typename Rhs>
struct binary_expression : expression<Lhs::arity + Rhs::arity>
{
    template <typename... Args>
    static constexpr decltype(auto) apply(Args&&... args)
    {
        static_assert(sizeof...(Args) == binary_expression::arity, "Wrong number of operands!");
        return _apply(std::make_index_sequence<Lhs::arity>{}, std::make_index_sequence<Rhs::arity>{}, std::tuple<Args&&...>(std::forward<Args>(args)...));
    }
    template <typename Tuple, std::size_t... Arity1, std::size_t... Arity2>
    static constexpr decltype(auto) _apply(std::index_sequence<Arity1...>, std::index_sequence<Arity2...>, Tuple&& args)
    {
        return Expr::eval(Lhs::apply(static_cast<typename std::tuple_element<Arity1, Tuple>::type>(std::get<Arity1>(args))...),
                          Rhs::apply(static_cast<typename std::tuple_element<Lhs::arity + Arity2, Tuple>::type>(std::get<Lhs::arity + Arity2>(args))...));
    }
};
struct arg : expression<1>
{
    template <typename Arg1>
    static constexpr decltype(auto) apply(Arg1&& arg1)
    {
        return std::forward<Arg1>(arg1);
    }
};
template <typename Type, Type value>
struct constant : expression<0>
{    
    static constexpr decltype(auto) apply()
    {
        return value;
    }
};
template <typename Rhs>
struct negate : unary_expression<negate<Rhs>, Rhs>
{
    template <typename Arg1>
    static constexpr decltype(auto) eval(Arg1&& arg1)
    {
        return -std::forward<Arg1>(arg1);
    }
};
template <typename Lhs, typename Rhs>
struct plus : binary_expression<plus<Lhs, Rhs>, Lhs, Rhs>
{
    template <typename Arg1, typename Arg2>
    static constexpr decltype(auto) eval(Arg1&& arg1, Arg2&& arg2)
    {
        return std::forward<Arg1>(arg1) + std::forward<Arg2>(arg2);
    }
};
template <typename Lhs, typename Rhs>
struct minus : binary_expression<minus<Lhs, Rhs>, Lhs, Rhs>
{
    template <typename Arg1, typename Arg2>
    static constexpr decltype(auto) eval(Arg1&& arg1, Arg2&& arg2)
    {
        return std::forward<Arg1>(arg1) - std::forward<Arg2>(arg2);
    }
};
template <typename Lhs, typename Rhs>
struct multiply : binary_expression<multiply<Lhs, Rhs>, Lhs, Rhs>
{
    template <typename Arg1, typename Arg2>
    static constexpr decltype(auto) eval(Arg1&& arg1, Arg2&& arg2)
    {
        return std::forward<Arg1>(arg1) * std::forward<Arg2>(arg2);
    }
};
int main()
{    
    // (1 + 2) + 3 = 6
    std::cout << plus<plus<arg, arg>, arg>::apply(1, 2, 3) << std::endl;
    // ((a + 5) + (2 * 6)) - 5 = 16
    int a = 4;
    std::cout << minus<plus<plus<arg, arg>, multiply<arg, constant<int, 6>>>, constant<int, 5>>::apply(a, 5, 2) << std::endl;
    // ((1 * 2) * 3) * 4 = 24
    std::cout << multiply<multiply<multiply<arg, arg>, arg>, arg>::apply(1, 2, 3, 4) << std::endl;
    // -((3 * 4) + (5 - 6)) = -11
    static_assert(negate<plus<multiply<arg, arg>, minus<arg, arg>>>::apply(3, 4, 5, 6) == -11, "!");
}

演示2

当然。这些被称为"表达模板"，你可以在这里找到SO的亮点。

早在90年代末，我就在POOMA系统上进行并行编程。不确定它是否已经更新到现代标准，但我看到它仍然可以在这里在线获得。POOMA的基础是一个名为PETE的"表达式模板引擎"，它可以重新用于其他评估引擎。此处描述了皮特。使用C++11，所有这些工作都会简单得多，我相信也有类似的工作使用这些更新的功能。

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