可变参数模板对象乘法 [英] variadic templated Object multiplication

问题描述

在下面的代码中，我正在为int值和对象做乘法可变参数模板.它适用于所有原始类型.它也仅适用于2个对象.但是，当我使用两个以上的自变量对象相乘时，代码不会编译.

In the below code I am doing multiplication variadic templates for the int values and also for the Objects. It works for the all primitive types. It also works for the only 2 objects. But the code doesn't compile when I use more than 2 arguments objects into multiply.

multiply(1, 2, 3, 4, 5, 6) //works correcyly
multiply( A(1), B(1))      //works correctly 
multiply( A(1), B(1), B(1) );   //compile time Error 
multiply( A(1), B(1), B(1), B(1) );   //compile time Error 

对于2个以上的对象乘法，如何解决此问题?乘法是按左联想完成的.

How could I solve this problem for the more than 2 object multiplication? Multiplication is done as left associative.

#include <iostream>
#include <assert.h>
#include <cstddef>
#include <typeinfo>
#include <stdlib.h>

using namespace std;

template <typename...> struct MulTs;

template <typename T1> struct MulTs<T1> {
    typedef T1 type;
};

template <typename T1, typename... Ts>
struct MulTs<T1, Ts...> {
    static typename MulTs < Ts...>::type makeTs(); //a
    static T1 makeT1(); //b
    typedef decltype(makeT1() * makeTs()) type; //c
};

template <typename T>
T multiply(const T& v) {
    return v;
}

template <typename T1, typename... Ts>
auto multiply(const T1& v1, const Ts&... rest) -> typename MulTs<T1,     Ts...>::type //instead of the decltype
{
    return v1 * multiply(rest...);
}

struct B;
struct A {
    friend A operator*(const A &, const B &);
    friend ostream & operator<<(ostream &os, const A &a);
    A(int val = 0) : i(val) {}
    private:
    const int i;
};
struct B {
    friend A operator*(const A &a, const B &b) {
        return A(a.i * b.i);
    }
    B(int val = 0) : i(val) {}
private:
    const int i;
};

ostream &operator<<(ostream &os, const A &a) {
    return os << a.i;
}

int main() {
    cout << multiply(1, 2, 3, 4, 5, 6) <<endl;//works correcyly
    cout << multiply( A(1), B(1))<<endl;      //works correctly 
    //cout << multiply( A(1), B(1), B(1) );   //compile time Error 
}

推荐答案

multiply( A(1), B(1), B(1) )

扩展为 A(1)*乘法(B(1)，B(1)).由于您没有重载 operator *(const B& ;, const B&)，因此会出现编译错误.

Expands to A(1) * multiply(B(1), B(1)). Since you don't have overloaded operator *(const B&, const B&) you're getting compile error.

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