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Generally speaking, it can't, because objects don't really have names. Essentially, assignment always binds a name to a value; the same is true of ``def`` and ``class`` statements, but in that case the value is a callable. Consider the following code::
Arguably the class has a name: even though it is bound to two names and invoked through the name ``B`` the created instance is still reported as an instance of class ``A``. However, it is impossible to say whether the instance's name is ``a`` or ``b``, since both names are bound to the same value.
Generally speaking it should not be necessary for your code to "know the names" of particular values. Unless you are deliberately writing introspective programs, this is usually an indication that a change of approach might be beneficial.
In comp.lang.python, Fredrik Lundh once gave an excellent analogy in answer to this question:
The same way as you get the name of that cat you found on your porch: the cat (object) itself cannot tell you its name, and it doesn't really care -- so the only way to find out what it's called is to ask all your neighbours (namespaces) if it's their cat (object)...
....and don't be surprised if you'll find that it's known by many names, or no name at all!
What's up with the comma operator's precedence?
Comma is not an operator in Python. Consider this session::
Since the comma is not an operator, but a separator between expressions the above is evaluated as if you had entered::
not::
The same is true of the various assignment operators (``=``, ``+=`` etc). They are not truly operators but syntactic delimiters in assignment statements.

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  1. Python documentation
  2. 3.12
  3. latest/faq/programming
  4. English
....and don't be surprised if you'll find that it's known by many names, or no name at all!
8 months ago
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read-only, rst-text
Source string location
../../faq/programming.rst:696
String age
8 months ago
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8 months ago
Source string age
8 months ago
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