为什么 unique_ptr 将删除器作为类型参数而 shared_ptr 没有？

Question

The std::unique_ptr template has two parameters: the type of the pointee, and the type of the deleter. This second parameter has a default value, so you usually just write something like std::unique_ptr<int> .

The std::shared_ptr template has only one parameter though: the type of the pointee. But you can use a custom deleter with this one too, even though the deleter type is not in the class template. The usual implementation uses type erasure techniques to do this.

Is there a reason the same idea was not used for std::unique_ptr ?

Answer 1

Part of the reason is that shared_ptr needs an explicit control block anyway for the ref count and sticking a deleter in isn't that big a deal on top. unique_ptr however doesn't require any additional overhead, and adding it would be unpopular- it's supposed to be a zero-overhead class. unique_ptr is supposed to be static.

You can always add your own type erasure on top if you want that behaviour- for example, you can have unique_ptr<T, std::function<void(T*)>> , something that I have done in the past.

Answer 2

Another reason, in addition to the one pointed out by DeadMG, would be that it's possible to write

std::unique_ptr<int[]> a(new int[100]);

and ~unique_ptr will call the correct version of delete (via default_delete<_Tp[]> ) thanks to specializing for both T and T[] .

Answer 3

From C++ Primer (5th Edition)<\/strong><\/em> , Chapter 16.1.6 - Efficiency and Flexibility