How Does Copy Assignment Handle R-Values When Move Is Deleted?

Imagine you have the following code snippet:

struct S
{
  S() = default(); 
  S(S const&){std::cout << "copy ctor" << '\n';};
  S& operator=(S const&) { std::cout << "copy assignment" << '\n'; return *this; }
  ~S();
};

S createS(){return s{};}

int main()
{
  S s{}; 

  s = createS(); 
}

Although it is against the rule of five, what happens in the assignment of s in the main() function? createS() returns an r-value. So, the compiler will search for a move assignment operator, and since it is not generated automatically and deleted, it will fall back to the copy assignment operator. However, an r-value semantically does not bind with a copy operation, since it does not have an identifier and cannot be reused (to my understanding).

My question is, how does the assignment operator handle this situation and distinguish between r-value and l-value?

My question is, how does the assignment operator handle this situation and distinguish between r-value and l-value?

It doesn't need to differentiate. Your copy assignment operator takes a const reference as input, and a const reference can bind to an rvalue (a non-const reference cannot). The temporary S object will not be destroyed until after the assignment is complete, so the reference is perfectly valid.