如果是左值，请参考;如果是右值，请进行复制，即使右值保持不变 [英] Take a reference if lvalue and make a copy if rvalue i.e. make rvalue persistent

问题描述

我在move和lvalue语义上还很新。我有一种错的印象。在这里
一旦实现 FunctContainer 我想要编写的代码：

I'm pretty new in move and lvalue semantics. And I have the impression I'm doing it wrong. Here the code I want to be able to write once FunctContainer is implemented:

std::function<double(double)> f = [](double x){return (x * x - 1); };

FunctContainer fc1 = FunctContainer(f);

FunctContainer fc2 = FunctContainer([](double x){return (x * x - 1); });

我想写 FunctContainer 的代码因此，存储在 fc1 中的函数的生存期是 f 之一，而生存期在 fc2 包含的功能是 fc2 本身的生存期。

I want to write FunctContainer's ctors so that the lifetime of the function stored in fc1 is the one of f and the lifetime in fc2 of the contained function is the lifetime of fc2 itself.

我写了一些东西（见下文），但我不太满意（我弄错了）。

I have written something (see below) but I'm not really satisfied (I got it wrong).

这是正确的c ++，但行为错误：当 f _ 是一个右值时， f _ 在调用构造函数后到期。

This is correct c++ but wrong behavior: f_ expires after the call to the constructor when f_ is an rvalue.

class FunctContainerWRONG{
public:
  IntegrandNotProjecting(const std::function<double(double)>& function)
    : f_(function){}
  IntegrandNotProjecting(const std::function<double(double)>&& function)
    : f_(std::move(function)){}
  const std::function<double(double)>& f_;
private:
};

这对我来说似乎很糟糕，可能不是正确的c ++，但旨在用伪代码解释所需的内容行为看起来像。如果可能，我想避免构造一个全新的对象，而只想让我的对象持久：

This looks awful at me and probably is not correct c++ but is intended to explain in pseudocode what the desired behavior looks like. If possible I want to avoid to constuct a brand new object and I just want to let my object "persist":

class FunctContainer{
public:
  FunctContainer(const std::function<double(double)>& function)
    : f_p(nullptr), 
      f_(function){}
  FunctContainer(const std::function<double(double)>&& function)
    : f_p()std::make_shared<std::function<double(double)>>(function)), 
      f_(*f_p){}
private:
    std::shared_ptr<std::function<double(double)>> f_p;
    const std::function<double(double)>& f_;
};

推荐答案

转发参考和引用折叠规则可以帮助您轻松实现这一目标。

Forwarding references and reference collapsing rules can help you achieve this easily.

template <typename T>
struct FunctContainer 
{
    T _f;

    template <typename TFwd>
    FunctContainer(TFwd&& f) : _f{std::forward<TFwd>(f)} { }
};

template <typename T>
auto makeFunctContainer(T&& f) -> FunctContainer<T>
{
    return {std::forward<T>(f)};
}

• 调用<$ c带有左值， T 的$ c> makeFunctContainer 将是左值引用。这意味着您将在 FucntContainer 内存储左值引用。

  • When you invoke makeFunctContainer with an lvalue, T will be an lvalue reference. This means that you'll store an lvalue reference inside FucntContainer.

    当您使用 rvalue 调用 makeFunctContainer 时， T 将是一个值。这意味着您将在 FucntContainer 中存储一个 value 。

    When you invoke makeFunctContainer with an rvalue, T will be a value. This means that you'll store a value inside FucntContainer.

    示例用法：

    auto fc0 = makeFunctContainer(std::move(f0)); // `fc0` owns `f0`
    auto fc1 = makeFunctContainer(f1); // `fc1` stores a reference to `f1`
    

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