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 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[].