c++ - 如何创建一个仿函数作为 2 个仿函数的总和？

Question

假设我有 2 个仿函数 add 和 mult。 add2(x) = x+2 和 mult3(x) = 3x。我想创建一个组合张量 t 使得 t(x) = x+2 + 3x。我提出了创建一个对 仿函数的解决方案，但我并不十分满意，因为该解决方案无法扩展到更多数量的仿函数。我有哪些选择？

struct add {
private:
    double c_;
public:
    add(double& c): c_(c) {}
    double operator()(double& x) {
        return c_ + x;
    }
};

struct mult {
private:
    double c_;
public:
    mult(double& c): c_(c) {}
    double operator()(double& x) {
        return c_ * x;
    }
}

template<typename T1, typename T2>
struct pair {
private:
    T1 t1_;
    T2 t2_;
public:
    pair(T1 t1, T2 t2): t1_(t1), t2_(t2) {}
    double operator()(double& x) {
        return t1_(x) + t2_(x);
    }
}

add add2 = add(2);
mult mult3 = mult(3);
pair combined = pair(add2, mult3);

Answer 1

名称 pair 并没有真正说明仿函数的作用。我认为按照以下思路创建一些东西是有意义的:

template<typename T1, typename T2>
struct plus
{
   plus(T1 t1, T2 t2) : t1_(t1), t2_(t2) {}
   double operator()(double x)
   {
      return t1_(x) + t2_(x);
   }
   T1 t1_;
   T2 t2_;
};

并提供一个辅助函数来创建这种类型的实例。

template<typename T1, typename T2>
plus<T1, T2> make_plus(T1 t1, T2 t2)
{
   return plus<T1, T2>(t1, t2);
}

然后，您可以使用:

add add2 = add(2);
mult mult3 = mult(3);
auto comb1 = make_plus(add2, mult3);
auto comb2 = make_plus(mult(10), make_plus(add2, mult3));

这是一个添加了一个减号的程序。

#include <iostream>

struct add
{
   add(double c): c_(c) {}
   double operator()(double x) {
      return c_ + x;
   }
   double c_;
};

struct mult
{
   mult(double c): c_(c) {}
   double operator()(double x) {
      return c_ * x;
   }
   double c_;
};

template<typename T1, typename T2>
struct plus
{
   plus(T1 t1, T2 t2) : t1_(t1), t2_(t2) {}
   double operator()(double x)
   {
      return t1_(x) + t2_(x);
   }
   T1 t1_;
   T2 t2_;
};

template<typename T1, typename T2>
plus<T1, T2> make_plus(T1 t1, T2 t2)
{
   return plus<T1, T2>(t1, t2);
}

template<typename T1, typename T2>
struct minus
{
   minus(T1 t1, T2 t2) : t1_(t1), t2_(t2) {}
   double operator()(double x)
   {
      return t1_(x) - t2_(x);
   }
   T1 t1_;
   T2 t2_;
};

template<typename T1, typename T2>
minus<T1, T2> make_minus(T1 t1, T2 t2)
{
   return minus<T1, T2>(t1, t2);
}


int main()
{
   add add2 = add(2);
   mult mult3 = mult(3);
   auto comb1 = make_plus(add2, mult3);
   auto comb2 = make_plus(mult(10), make_plus(add2, mult3));
   auto comb3 = make_minus(mult(10), make_plus(add2, mult3));

   std::cout << comb1(10) << std::endl;
   std::cout << comb2(10) << std::endl;
   std::cout << comb3(10) << std::endl;
}

输出:

42
142
58

---

另一种选择是在特定应用程序中创建类型 add、mult、plus 和 minus命名空间，然后添加 operator+ 和 operator- 重载，使 main 中的代码更加直观。

#include <iostream>

namespace myapp
{
   struct identity
   {
      double operator()(double x) { return x;}
   };

   struct add
   {
      add(double c): c_(c) {}
      double operator()(double x) {
         return c_ + x;
      }
      double c_;
   };

   struct mult
   {
      mult(double c): c_(c) {}
      double operator()(double x) {
         return c_ * x;
      }
      double c_;
   };

   template<typename T1, typename T2>
      struct plus
      {
         plus(T1 t1, T2 t2) : t1_(t1), t2_(t2) {}
         double operator()(double x)
         {
            return t1_(x) + t2_(x);
         }
         T1 t1_;
         T2 t2_;
      };

   template<typename T1, typename T2>
      plus<T1, T2> make_plus(T1 t1, T2 t2)
      {
         return plus<T1, T2>(t1, t2);
      }

   template<typename T1, typename T2>
      plus<T1, T2> operator+(T1 t1, T2 t2)
      {
         return make_plus(t1, t2);
      }

   template<typename T1, typename T2>
      struct minus
      {
         minus(T1 t1, T2 t2) : t1_(t1), t2_(t2) {}
         double operator()(double x)
         {
            return t1_(x) - t2_(x);
         }
         T1 t1_;
         T2 t2_;
      };

   template<typename T1, typename T2>
      minus<T1, T2> make_minus(T1 t1, T2 t2)
      {
         return minus<T1, T2>(t1, t2);
      }

   template<typename T1, typename T2>
      minus<T1, T2> operator-(T1 t1, T2 t2)
      {
         return make_minus(t1, t2);
      }
}

int main()
{
   using namespace myapp;
   add add2 = add(2);
   mult mult3 = mult(3);

   auto comb1 = add2 + mult3;
   auto comb2 = mult(10) +  add2 + mult3;
   auto comb3 = mult(10) - (add2 + mult3);
   auto comb4 = identity{} + add(25);

   std::cout << comb1(10) << std::endl;
   std::cout << comb2(10) << std::endl;
   std::cout << comb3(10) << std::endl;
   std::cout << comb4(10) << std::endl;
}