指南做constexpr操作符重载？

问题描述

考虑一个简单的 Wrapper 带有重载乘法的类 operator * = 和 。对于旧式运算符重载，可以按运算符* = 定义运算符* 甚至是等图书馆及其通过@DanielFrey的来减少你的样板。

但是，对于使用新的C ++ 11 constexpr 的编译时计算，这种方便就消失了。由于后者修改其（隐式）左参数，因此 constexpr operator * 不能调用 operator * = 。此外，还有，因此添加了一个额外的 constexpr运算符* 到现有的运算符* 会导致重载解析模糊。

我目前的方法是：

  #include< iostream> 
 
 struct Wrap 
 {
 int value; 
 
 Wrap& operator * =（Wrap const& rhs）
 {value * = rhs.value; return * this; } 
 
 //需要注释此函数，因为使用constexpr版本重载模糊性
 // friend Wrap运算符*（Wrap const& lhs，Wrap const& rhs）
 / / {return Wrap {lhs} * = rhs; } 
 
 friend constexpr Wrap运算符*（Wrap const& lhs，Wrap const& rhs）
 {return {lhs.value * rhs.value}; } 
}; 
 
 constexpr Wrap factorial（int n）
 {
 return n？ factorial（n-1）* Wrap {n}：Wrap {1}; 
} 
 
 //希望能够静态初始化这些数组
 struct Hold 
 {
 static constexpr Wrap Int [] = {factorial ），factorial（1），factorial（2），factorial（3）}; 
}; 
 
 int main（）
 {
 std :: cout< Hold :: Int [3] .value<< \\\
; // 6 
 auto w = Wrap {2}; 
 w * = Wrap {3}; 
 std :: cout<< w.value<< \\\
; // 6 
} 
  

。我的问题是：

• 运算符* = 运算符* ，而不是运算符* / code>


• 因此，Boost.Operators不再适用于减少编写许多其他算术运算符的样板。

问题：这是推荐的C ++ 11方式，同时具有运行时 operator * = 混合运行时/编译时 constexpr operator * ？ C ++ 14改变这里的任何东西。减少逻辑重复？

UPDATE ：@AndyProwl的答案被接受为惯用的，但根据@DyP的建议，在C + 11一个 可以减少逻辑重复，但需要额外的分配和反直觉风格。

  // define operator * = in operator of operator * 
 Wrap& operator * =（Wrap const& rhs）
 {* this = * this * rhs; return * this; } 
  

解决方案我找不到一个惯用的解决方案C + +11（虽然作为解决方法，似乎可以接受我）。

在C ++ 14中，（见C ++ 14标准草案n3690的附录C.3.1）可以简单地定义 operator * = 和 operator * 为 constexpr

  struct Wrap 
 {
 int value; 
 
 constexpr Wrap& operator * =（Wrap const& rhs）
 {value * = rhs.value; return * this; } 
 
 friend constexpr Wrap operator *（Wrap const& lhs，Wrap const& rhs）
 {return Wrap（lhs）* = rhs; } 
}; 
  

这是一个，其中上述程序使用 -std = c ++ 1y on Clang - 不幸的是，GCC似乎没有实施这条规则。

Consider a simple int Wrapper class with overloaded multiplication operator*= and operator*. For "old-style" operator-overloading, one can define operator* in terms of operator*=, and there are even libraries like Boost.Operators and its modern incarnation df.operators by @DanielFrey that reduce the boilerplate for you.

However, for compile-time computations using the new C++11 constexpr, this convenience disappears. A constexpr operator* cannot call operator*= because the latter modifies its (implicit) left argument. Furthermore, there is no overloading on constexpr, so adding an extra constexpr operator* to the existing operator* results in an overload resolution ambiguity.

My current approach is:

#include <iostream>

struct Wrap
{
    int value;

    Wrap& operator*=(Wrap const& rhs)
    { value *= rhs.value; return *this; }

    // need to comment this function because of overloading ambiguity with the constexpr version
    // friend Wrap operator*(Wrap const& lhs, Wrap const& rhs)
    // { return Wrap { lhs } *= rhs; }

    friend constexpr Wrap operator*(Wrap const& lhs, Wrap const& rhs)
    { return { lhs.value * rhs.value }; }
};

constexpr Wrap factorial(int n)
{
    return n? factorial(n - 1) * Wrap { n } : Wrap { 1 };
}

// want to be able to statically initialize these arrays
struct Hold
{
    static constexpr Wrap Int[] = { factorial(0), factorial(1), factorial(2), factorial(3) };
};

int main()
{
    std::cout << Hold::Int[3].value << "\n"; // 6
    auto w = Wrap { 2 };
    w *= Wrap { 3 };
    std::cout << w.value << "\n"; // 6
}

Live output here. My problems with this are:

• duplication of the multiplication logic in both operator*= and operator*, instead of operator* being expressed in terms of operator*=
• hence, Boost.Operators no longer works to reduce the boilerplate for writing many other arithmetic operators

Question: is this the recommended C++11 way of having both a run-time operator*= and mixed run-time/compile-time constexpr operator*? Does C++14 change anything here to e.g. reduce the logic duplication?

UPDATE: The answer by @AndyProwl is accepted as idiomatic but as per suggestion of @DyP, in C++11 one could reduce the logic duplication at the expense of an extra assignment and counter-intuitive style

    // define operator*= in terms of operator*
    Wrap& operator*=(Wrap const& rhs)
    { *this = *this * rhs; return *this; }

解决方案

I could not find an idiomatic solution for C++11 (although as a workaround, DyP's suggestion seems acceptable to me).

In C++14 however, where constexpr does not imply const (see Annex C.3.1 of the C++14 Standard Draft n3690), you could simply define both operator *= and operator * as constexpr, and define the latter in terms of the former, as usual:

struct Wrap
{
    int value;

    constexpr Wrap& operator *= (Wrap const& rhs)
    { value *= rhs.value; return *this; }

    friend constexpr Wrap operator * (Wrap const& lhs, Wrap const& rhs)
    { return Wrap(lhs) *= rhs; }
};

Here is a live example, where the above program is being compiled with -std=c++1y on Clang - unfortunately, GCC does not seem to implement this rule yet.

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