问题
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?
回答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.
回答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[].
来源：https://stackoverflow.com/questions/6829576/why-does-unique-ptr-have-the-deleter-as-a-type-parameter-while-shared-ptr-doesn