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Home/ Questions/Q 9195723
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T21:41:28+00:00 2026-06-17T21:41:28+00:00

Java has Arrays.fill(A,1) . For a pre-existing List A , is there a shortcut

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Java has Arrays.fill(A,1). For a pre-existing List A, is there a shortcut for filling the list with 1? I am writing a function that takes an array and changes the array in some ways. Since arrays are pointers, my function will not return an array. The caller will see the changes after my function returns. The first step in my function is to fill the array with 1s. Doing

def my_work(A):
   A =[1]*len(A) 
   # more work on A

does not seem to change A when my_work is done.

So is my only option

for i in range(len(A)):
  A[i]=1

or is there a shortcut? Mine looks like a workaround.

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  1. Editorial Team
    Editorial Team
    2026-06-17T21:41:29+00:00Added an answer on June 17, 2026 at 9:41 pm

    If you really want to change A in place, the A[:] syntax should work:

    >>> A = [1,2,3]
    >>> def my_work(A):
    ...     A[:] = [1]*len(A) 
    ...     
    >>> A
    [1, 2, 3]
    >>> my_work(A)
    >>> A
    [1, 1, 1]
    

    And here is the relevant section of the tutorial (“assignment to slices”).

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