避免在构造函数中const引用和rvalue引用呈指数增长 [英] Avoid exponential grow of const references and rvalue references in constructor

问题描述

我正在为机器学习库编写一些模板化类，并且我经常面对这个问题。我主要使用策略模式，其中类接收不同功能的模板参数策略，例如：

I am coding some templated classes for a machine learning library, and I'm facing this issue a lot of times. I'm using mostly the policy pattern, where classes receive as template argument policies for different functionalities, for example:

template <class Loss, class Optimizer> class LinearClassifier { ... }

问题出在构造函数上。随着策略（模板参数）数量的增长，const引用和rvalue引用的组合呈指数增长。在上一个示例中：

The problem is with the constructors. As the amount of policies (template parameters) grows, the combinations of const references and rvalue references grow exponentially. In the previous example:

LinearClassifier(const Loss& loss, const Optimizer& optimizer) : _loss(loss), _optimizer(optimizer) {}

LinearClassifier(Loss&& loss, const Optimizer& optimizer) : _loss(std::move(loss)), _optimizer(optimizer) {}

LinearClassifier(const Loss& loss, Optimizer&& optimizer) : _loss(loss), _optimizer(std::move(optimizer)) {}

LinearClassifier(Loss&& loss, Optimizer&& optimizer) : _loss(std::move(loss)), _optimizer(std::move(optimizer)) {}

有什么办法可以避免这种情况？

Is there some way to avoid this?

推荐答案

实际上，这就是 完美转发 。将构造函数重写为

Actually, this is the precise reason why perfect forwarding was introduced. Rewrite the constructor as

template <typename L, typename O>
LinearClassifier(L && loss, O && optimizer)
    : _loss(std::forward<L>(loss))
    , _optimizer(std::forward<O>(optimizer))
{}

但这可能会更简单Ilya Popov在他的答案中的建议。老实说，我通常是这样做的，因为移动的目的是廉价的，再移动一次并不会显着改变事情。

But it will probably be much simpler to do what Ilya Popov suggests in his answer. To be honest, I usually do it this way, since moves are intended to be cheap and one more move does not change things dramatically.

霍华德·辛南特 https://stackoverflow.com/questions/36868442/avoid-exponential-grow-of-const-references-and-rvalue-references-in-constructor/36868542#comment61305289_36868542>我的方法可以是SFINAE不友好的，因为现在LinearClassifier接受构造函数中的任何一对类型。 巴里的答案显示了如何处理它。

As Howard Hinnant has told, my method can be SFINAE-unfriendly, since now LinearClassifier accepts any pair of types in constructor. Barry's answer shows how to deal with it.

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