what does "unqualified on right hand side" mean in OOPs (Python)?
I came across "unqualified on right hand side" phrase while reading oops concept in python for usage like self._customer = customer. What that phrase trying to explain?
Complete statement is
For example, the command, self._customer = customer, assigns the instance variable self._customer to the parameter customer; note that because customer is unqualified on the right-hand side, it refers to the parameter in the local namespace. --Data Structures and Algorithms in Python p. 72
A dotted name showing the “path” from a module’s global scope to a class, function or method defined in that module, as defined in PEP 3155. For top-level functions and classes, the qualified name is the same as the object’s name:
When used to refer to modules, the fully qualified name means the entire dotted path to the module, including any parent packages, e.g. email.mime.text:
Put more simply, qualifying a name in Python means that you explicitly define its scope. Thus self._customer is a qualified name (it identifies the instance variable customer for the enclosing class) whereas the bare customerreference does not specify any scope qualifications.
When a name is unqualified, Python applies Lexical Scoping rules to try and find the variable, searching (in order)
Local variables (including function parameters)
Variables local to any outer functions, if we're dealing with a nested function definition
Asked in February 2016Viewed 3,102 timesVoted 14Answered 1 times