Why type is not required for smart pointer of type shared pointer?

Question

Hi

AFIK, shared_ptr is a template class. To create object of that class , we need to specify type.

Refer attached code and observe below line into print function:

shared_ptr temp = head;

why type node is not required like shared_ptr<node> ? I was expecting compiler error without type but it works fine..

How this is possible?
https://code.sololearn.com/crYV1fiFiNPr/?ref=app

Answer 1

Short answer: C++17.
Longer answer:
https://en.cppreference.com/w/cpp/language/class_template_argument_deduction