Python Var Scope Fix
Excerpt
A conversational AI system that listens, learns, and challenges
In Python, variables are passed by object reference. This means that when you pass a variable to a function, you’re actually passing a reference to the object the variable points to, not a copy of the object itself.
However, this can sometimes lead to confusion because of how mutable and immutable objects behave. Immutable objects like integers, strings, and tuples cannot be modified in place, so any attempt to modify them within a function creates a new object. Mutable objects like lists and dictionaries, on the other hand, can be modified in place.
Here’s a quick example to illustrate this behavior:
<div><p><code id="code-lang-python"><span>def</span> <span>modify_list</span>(<span>lst</span>):
lst.append(<span>4</span>)
my_list = [<span>1</span>, <span>2</span>, <span>3</span>]
modify_list(my_list)
<span>print</span>(my_list) <span># Output: [1, 2, 3, 4]</span>
</code></p></div>
In the example above, even though my_list is passed to the function modify_list, and the list is modified within the function, the changes are reflected outside the function. This is because lists are mutable, and changes made to them affect the original object.
However, if you try to reassign the parameter inside the function, it won’t affect the original variable:
<div><p><code id="code-lang-python"><span>def</span> <span>reassign</span>(<span>lst</span>):
lst = [<span>4</span>, <span>5</span>, <span>6</span>]
my_list = [<span>1</span>, <span>2</span>, <span>3</span>]
reassign(my_list)
<span>print</span>(my_list) <span># Output: [1, 2, 3]</span>
</code></p></div>
In this case, lst inside the function reassign is reassigned to a new list, but this change does not affect the original my_list variable outside the function. This behavior is because the parameter lst is a new reference to a different object, not a modification of the original list.