Sum and Product monoid in C++

Monoid.cpp

#include <iostream>
#include <cstdlib>
#include <memory>
#include <vector>
#include <string>

template<typename M, typename T>
class SemiGroup {
protected:
    T m_t;
public:
    SemiGroup(T t) : m_t(t) {}
    SemiGroup<M,T> mappend(SemiGroup<M,T> other);

    T get() { return m_t; }
};

template<typename M, typename T>
class Monoid : public SemiGroup<M,T> {
public:
    Monoid(T t) : SemiGroup<M,T>(t) {}
    static Monoid<M,T> mzero;
};

class Sum {};

template<>
SemiGroup<Sum, int> SemiGroup<Sum, int>::mappend(SemiGroup<Sum, int> other) {
    return SemiGroup<Sum, int>(m_t + other.m_t);
}

template<>
Monoid<Sum, int> Monoid<Sum, int>::mzero = Monoid<Sum, int>(0);

class Product {};

template<>
SemiGroup<Product, int> SemiGroup<Product, int>::mappend(SemiGroup<Product, int> other) {
    return SemiGroup<Product, int>(m_t * other.m_t);
}

template<>
Monoid<Product, int> Monoid<Product, int>::mzero = Monoid<Product, int>(1);

template<typename M, typename T>
Monoid<M, T> mconcat(const std::vector<T> & vals) {
    Monoid<M, T> r = Monoid<M, T>::mzero;
    for (auto v : vals) {
        r = Monoid<M, T>(r.mappend(v).get());
    }
    return r;
}

int main(void) {
    std::vector<int> vals{3, 5};
    std::cout << "Sum::mconcat [3,5] == "
              << mconcat<Sum,int>(vals).get()
              << std::endl;
    std::cout << "Product::mconcat [3,5] == "
              << mconcat<Product,int>(vals).get()
              << std::endl;
    return 0;
}

（心悦诚服（但是你的写法比我的长啊））@darkfall

(睡前意识模糊状态下的, 其实没啥卵区别, 除了variadic template可以传任意多个参数进去而不用构造list, 几个模板单独出来个人觉得扩展比较方便, 这样就不用偏特化你的类而是几个trait模板(不用在意那些细节( @wecing