Why doesn't returning a std::unique_ptr violate unique_ptr's deleted copy constructor? [duplicate]

I'm learning about copy constructors; originally started with trying to understand why I got a "use of deleted function" error when I had a unique_ptr as an argument in a class' constructor. That led me to learn that unique_ptrs are intentionally not copyable.

But then why does the following not violate non-copyability of unique_ptrs? Function foo() returns a unique_ptr - so a copy is placed on the stack then assigned to variable p. I'm not following something here, because this looks like the copy constructor should be invoked.

// main.cpp
#include <memory>
#include <iostream>

std::unique_ptr<int> foo() {
  return std::unique_ptr<int>(new int( 42 ));
}

int main( int argc, char* argv[] ) {
  std::unique_ptr<int> p = foo();

  std::cout << *p << std::endl;

  return 0;
}

$ g++ --version && g++ -g ./main.cpp && ./a.out
g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

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