Home >Backend Development >Python Tutorial >Why Does Assigning a New List to a Function Argument Not Modify the Original List in Python?
In Python, when a function receives a list as an argument, the parameter effectively becomes a reference pointing to the original list's location in memory. Any modifications made to the list within the function are directly reflected in the calling list.
However, the following code demonstrates a peculiar behavior that challenges this understanding:
<code class="python">def function1(list_arg): a = function2() # function2 returns an array of numbers list_arg = list(a) list1 = [0] * 5 function1(list1) print(list1) # [0, 0, 0, 0, 0]</code>
In this example, we intend for function1 to modify list1 by replacing its elements with those from array a. But the result shows list1 remains unchanged. Why does this happen?
The issue arises from the incorrect assumption that assigning a new value to the list_arg variable within function1 modifies the original list. In reality, this assignment only changes the reference of list_arg to point to a different object in memory. The original list referenced by the calling code remains untouched.
To successfully alter the original list within function1, we need to modify its individual elements. This can be accomplished through slice assignment:
<code class="python">list_arg[:] = list(a)</code>
Slice assignment replaces the entire range of elements in list_arg with the values from list(a). Effectively, this action directly modifies the original list.
While slice assignment provides a convenient solution, it is important to note its potential for confusion. By modifying a list in place, the code becomes less transparent and more susceptible to misunderstandings by other developers maintaining it. Consider alternative approaches, such as returning a new modified list from the function, to enhance code clarity and readability.
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