How does Python's name and object reference model work?

Q: How does Python's name and object reference model work?

**Target Audience:** Python beginners and intermediate developers preparing for interviews — you only need to know basic assignment and lists. **Difficulty:** Medium **Python version:** Behaviour checked on CPython 3.11, 3.12 and 3.13; the examples run in the page on Pyodide 0.25.1 (CPython 3.11.3). > **How to read this doc:** Each idea is explained in plain language first, then tagged with `📌 Interview term:` so you also learn the exact word an interviewer expects. Every output shown was produced by actually running the code — see the **Verified** section. ## 1. Why This Even Matters — A Story First In many languages a variable is a **box**: `x = 5` puts the number 5 into a box called `x`, and `y = x` copies the 5 into a second box. Python does not work like that. In Python, values are **balloons** floating around, and names are **luggage tags on strings** tied to balloons. `a = [1, 2]` creates a balloon and ties a tag `a` to it. `b = a` does not make a new balloon; it ties a seco

The problem

Question presented to candidate: "People say Python has no variables, only names. What does that mean? Explain what happens in memory when you write a = [1, 2] and then b = a."

What a strong answer should cover:

Every value is an object with an identity, a type and a value; a name is a label in a namespace that refers to an object. Assignment binds a name and never copies.
Aliasing: two names can refer to one object, so a mutation through one is visible through the other; is and id() compare identity.
Rebinding a name (a = a + [3]) vs mutating the object (a.append(3)); += mutates lists in place but rebinds for immutable types like int and tuple.
Function calls bind parameters to the caller's objects ("call by sharing"), so mutation is visible to the caller and rebinding is not.
del removes a name, not an object; CPython frees the object when its reference count reaches zero, and sys.getrefcount shows the count (with caveats such as immortal objects in 3.12+).
Classic traps: [[0] * 3] * 2 shares rows, and t[0] += [x] on a tuple both mutates the list and raises TypeError.

Code / implementation expected: Optional — predict output of an aliasing snippet, using is or id() to explain it.

memory
references

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Solution

Target Audience: Python beginners and intermediate developers preparing for interviews — you only need to know basic assignment and lists. Difficulty: Medium Python version: Behaviour checked on CPython 3.11, 3.12 and 3.13; the examples run in the page on Pyodide 0.25.1 (CPython 3.11.3).

How to read this doc: Each idea is explained in plain language first, then tagged with 📌 Interview term: so you also learn the exact word an interviewer expects. Every output shown was produced by actually running the code — see the Verified section.

1. Why This Even Matters — A Story First

In many languages a variable is a box: x = 5 puts the number 5 into a box called x, and y = x copies the 5 into a second box. Python does not work like that. In Python, values are balloons floating around, and names are luggage tags on strings tied to balloons. `a = [1,

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10 min read · 12 sections

Run the code

Edit the code and run it here. Each console line with an expected value becomes a test.

PythonNames, objects and aliasing: rebinding vs mutation

Practice

PythonThe classic traps, reference counts and when objects die

Practice

Discussion

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