为什么unique_ptr需要两个模板参数，当shared_ptr只需要一个？ [英] Why does unique_ptr take two template parameters when shared_ptr only takes one?

问题描述

unique_ptr 和 shared_ptr 接受一个自定义析构函数来调用他们自己的对象。但是在 unique_ptr 的情况下，析构函数作为类的模板参数传递，其中 shared_ptr

Both unique_ptr and shared_ptr accept a custom destructor to call on the object they own. But in the case of unique_ptr, the destructor is passed as a template parameter of the class, wherease the type of shared_ptr's custom destructor is to be specified as a template parameter of the constructor.

template <class T, class D = default_delete<T>> 
class unique_ptr
{
    unique_ptr(T*, D&); //simplified
    ...
};

和

template<class T>
class shared_ptr
{
    template<typename D>
    shared_ptr(T*, D); //simplified
    ...
};

我看不出为什么会有这样的差别。

I can’t see why such difference. What requires that?

推荐答案

如果你提供的删除作为模板参数（如 unique_ptr ）它是类型的一部分，你不需要在这个类型的对象中存储任何额外的东西。
如果deleter作为构造函数的参数传递（如 shared_ptr ），则需要将其存储在对象中。这是额外的灵活性的代价，因为你可以为同一类型的对象使用不同的删除器。

If you provide the deleter as template argument (as in unique_ptr) it is part of the type and you don't need to store anything additional in the objects of this type. If deleter is passed as constructor's argument (as in shared_ptr) you need to store it in the object. This is the cost of additional flexibility, since you can use different deleters for the objects of the same type.

我想这是原因： unique_ptr 应该是非常轻量级的对象，零开销。用每个 unique_ptr 存储删除器可以使其大小增加一倍。因为人们会使用好的旧原始指针，而这是错误的。

I guess this is the reason: unique_ptr is supposed to be very lightweight object with zero overhead. Storing deleters with each unique_ptr could double their size. Because of that people would use good old raw pointers instead, which would be wrong.

另一方面， shared_ptr 不是那么轻量级，因为它需要存储引用计数，所以存储自定义删除程序看起来好像是好的交换。

On the other hand, shared_ptr is not that lightweight, since it needs to store reference count, so storing a custom deleter too looks like good trade off.

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