Why does `list_arg = list(a)` not modify the original list in `function1` in Python?

Modifying Lists as Function Parameters in Python

Consider the following code:

<code class="python">def function1(list_arg):
    a = function2()           # function2 returns an array of numbers
    list_arg = list(a)       # creates a new list and assigns it to list_arg

list1 = [0] * 5
function1(list1)
print(list1)      # Output: [0, 0, 0, 0, 0]</code>

This code attempts to modify the list passed as a parameter to function1. However, if we print the list after calling function1, we find that it remains unchanged, despite the assignment within the function. This is because assigning a new list to list_arg within the function breaks the reference to the original list.

To modify the elements of the list in place, we can use assignment slicing:

<code class="python">list_arg[:] = list(a)</code>

This line assigns the elements of the new list created from a to the slice of list_arg that starts from the beginning and goes to the end. This effectively replaces the elements of list_arg with the elements of a.

Caution:

In-place list modifications can be confusing. It's generally preferred to create a new list with the desired modifications and assign it to the parameter variable, as this preserves a clear separation between the original list and the modified list.

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