什么是“最佳"?现代C ++中重载算术运算符的方法? [英] What is the "best" way to overload arithmetic operators in modern C++?
问题描述
我想在C ++中实现一种类似于数字"的数学对象(例如，组或环中的元素).我确信我不是唯一一个处理此问题的人，因此可能有很多关于什么是重载算术运算符的最佳"方法的讨论.但是，我找不到满意的答案(尽管也许我太懒了以至于无法谷歌搜索更多内容.)
I want to implement a kind of "number-like" mathematical objects (say, elements in a group or a ring) in C++. I'm sure that I'm not the only one dealing with this problem, so there may be abundant amount of discussions about what is the "best" way of overloading arithmetic operators. However, I couldn't find satisfactory answers (although maybe I was too lazy to googling more).
假设我要为A类重载运算符"+". 问题是，我能想到太多不同的重载:
Let's say I want to overload the operator "+" for a class A. The problem is, there are too many different overloads I can think of:
- operator +(const A& x，const A& y)
- operator +(const A& x，A&&y)
- operator +(A&& x，const A& y)
- operator +(A& x，A& y)
- A :: operator + =(const A& y)&
- A :: operator + =(const A& y)&&
- A :: operator + =(A& y)&
- A :: operator + =(A& y)&&
第一个问题.为了使对A实例的操作尽可能高效且尽可能像原始类型一样"，是否需要所有这些重载?我认为在普通"情况下，不需要(或不应)重载右值限定的A :: operator + =.这样对吗?我认为所有1-4都是必要的.例如，对于情况3，由于x被移动，我们不需要分配新的空间来保存返回值，并且我们可以安全地重用为x保留的分配存储.是吗?
First question. Do all of these overloads necessary to make operations on instances of A as efficient as, and as much as "being like primitive types" as possible? I think for "ordinary" cases, rvalue-qualified A::operator+= need not (or should not) be overloaded. Is it right? I think all of 1-4 are necessary. For example, for the case 3, since x is being moved, we do not need to allocate a new space to hold the return value, and we can safely reuse the allocation storage reserved for x. Is it right?
第二个问题.我认为在大多数情况下，所有这些重载都可能共享许多代码.如何在不牺牲性能等的情况下最小化代码重复?是否有任何特殊的技巧/习惯来做到这一点? 我更喜欢通用和可扩展的方法(如果存在).
Second question. I think all of these overloads may share a lot of codes for most of this kind of cases. How can I minimize the code duplication without sacrificing performance/etc.? Is there any special techniques/idioms to do this? I prefer general and extensible methods, if exist.
推荐答案
不可能给出最笼统的答案，因为选择最佳"方式将涉及对操作细节的理解.
It's not possible to give a most-general answer, because choosing the "best" way will involve an appreciation of details about the operation.
例如，最常见的模式(我在下面给出)对矩阵乘法不太好，因为在那种情况下，最简单的方法是声明第一个从零开始并读取两个参数的矩阵.另外，在这种情况下，您可能都想使用惰性评估.
For example the most common pattern (which I give below) is not so good for Matrix Multiplication, because in that case it is simplest to declare a third matrix that starts off zeroed and reads the two arguments. Also you may want to use lazy evaluation in which case none of this applies.
我建议确保您的代码在所有情况下都能给出正确的答案，一旦程序运行并且您对语言有更多的经验，以后就可以担心微优化.
I would recommend making sure that your code gives the correct answers for all cases, and you can worry about micro-optimization later once your program works and once you have more experience in the language.
对于实现+的最有效方法是修改参数之一的类，则以下两种情况涵盖了所有用法，并具有强烈的异常保证:
For a class where the most efficient way to implement + is to modify one of the arguments, then the following two cases cover all uses, with the strong exception guarantee:
A& A::operator+=(A const &y) { /* modify *this using y */ ; return *this; }
A operator+ ( A x, A const& y ) { x += y; return x; }
有关上述代码的作用以及原因的更多说明，请参见运算符重载大线程.
For more explanation about what the above code does and why, see the operator overloading megathread.
在C ++ 03中，使用A const& x代替A x并没有多大区别，但是在C ++ 11中，对于第一个参数是右值的情况，这稍为最佳，因为资源现在可以从第一个参数中偷走了.
In C++03 it didn't make much difference to use A const& x instead of A x, but in C++11 this is slightly more optimal for the case where the first argument is an rvalue, because resources can now be stolen from the first argument.
关于在operator+=上使用ref限定符的决定. &和&&单独重载没有任何好处.如果您看到人们在使用&，则其理由不是也要重载，而是要为在右值上使用+=的尝试给出编译错误；否则，可能会出现编译错误.理由是这很可能是错误的.
Regarding the decision to use ref-qualifiers on operator+=. There is no benefit to overloading separate for & and &&. If you have seen people using &, the rationale is not to also overload, but to give a compile error for the attempt to use += on an rvalue; with the rationale being that that's likely to be a mistake.
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