does auto deduce that a type is a pointer?

Question

I am unclear about the usage of auto in a function I am looking over. Given a parameter defined like:

someFunction(const unique_ptr<BSTNode<int>>& node, paramType param) {

the body of the function establishes a local variable like this:

auto *localNode = node.get();

My understanding of auto is that it deduces the type of what is assigned to it at compile time. The docs state that unique_ptr::get() returns a pointer to the object that it manages. Given that fact, why is it necessary to add * to the variable declaration?

Answer 1

Using a * to deduce a pointer is unnecessary, but may be desirable. If the code is later changed such that the initializer no longer returns a pointer, the * will cause auto deduction to fail. If the code that uses this variable is designed with a pointer in mind, this may be a good thing.

Answer 2

why is it necessary to add * to the variable declaration?

It's not. The benefit is that it makes it exceedingly obvious to the reader that localNode is a pointer.