Why does unique_ptr have the deleter as a type parameter while shared_ptr doesn't?

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 somet

Answer 1

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 2

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.