为了支持移动语义，函数参数应该由unique_ptr，按值还是按rvalue接受? [英] To support move semantics, should function parameters be taken by unique_ptr, by value, or by rvalue?

问题描述

我的函数之一将向量作为参数并将其存储为成员变量.我正在使用const引用向量，如下所述.

One of my function takes a vector as a parameter and stores it as a member variable. I am using const reference to a vector as described below.

class Test {
 public:
  void someFunction(const std::vector<string>& items) {
   m_items = items;
  }

 private:
  std::vector<string> m_items;
};

但是，有时items包含大量字符串，因此我想添加一个支持移动语义的函数(或用新函数替换该函数).

However, sometimes items contains a large number of strings, so I'd like to add a function (or replace the function with a new one) that supports move semantics.

我正在考虑几种方法，但是我不确定该选择哪种方法.

I am thinking of several approaches, but I'm not sure which one to choose.

1)unique_ptr

1) unique_ptr

void someFunction(std::unique_ptr<std::vector<string>> items) {
   // Also, make `m_itmes` std::unique_ptr<std::vector<string>>
   m_items = std::move(items);
}

2)按值传递并移动

void someFunction(std::vector<string> items) {
   m_items = std::move(items);
}

3)右值

void someFunction(std::vector<string>&& items) {
   m_items = std::move(items);
}

我应该避免使用哪种方法，为什么?

Which approach should I avoid and why?

推荐答案

除非您有理由让vector驻留在堆中，否则我建议不要使用unique_ptr

Unless you have a reason for the vector to live on the heap, I would advise against using unique_ptr

向量的内部存储无论如何都驻留在堆上，因此，如果使用unique_ptr，则将需要2个间接度，一个将引用该向量的指针取消引用，然后再次解引用内部存储缓冲区.

The vector's internal storage lives on the heap anyway, so you'll be requiring 2 degrees of indirection if you use unique_ptr, one to dereference the pointer to the vector, and again to dereference the internal storage buffer.

因此，我建议使用2或3.

As such, I would advise to use either 2 or 3.

如果使用选项3(需要右值引用)，则在调用.

If you go with option 3 (requiring an rvalue reference), you are foisting a requirement on the users of your class that they pass an rvalue (either directly from a temporary, or move from an lvalue), when calling someFunction.

从左值移出的要求很繁琐.

The requirement of moving from an lvalue is onerous.

如果您的用户想要保留矢量的副本，则必须跳过这一步.

If your users want to keep a copy of the vector, they have to jump through hoops to do so.

std::vector<string> items = { "1", "2", "3" };
Test t;
std::vector<string> copy = items; // have to copy first
t.someFunction(std::move(items));

但是，如果您选择选项2，则用户可以决定是否要保留副本-选择权是他们的

However, if you go with option 2, the user can decide if they want to keep a copy, or not - the choice is theirs

保留副本:

std::vector<string> items = { "1", "2", "3" };
Test t;
t.someFunction(items); // pass items directly - we keep a copy

请勿保留副本:

std::vector<string> items = { "1", "2", "3" };
Test t;
t.someFunction(std::move(items)); // move items - we don't keep a copy

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