Learning python: Why these two function yield different results?

Asked Oct 20 '21 at 19:10

Active Oct 20 '21 at 19:10

Viewed 23 times

def fun1(a):
  a=[7, 8]

def fun2(a):
  a[0]=7
  a[1]=8


q=[1,2]

fun1(q)
print(q)

fun2(q)
print(q)

The first one prints out: [1, 2] which means that q has not been changed using fun1. The second one prints out: [1, 2] which means that q is changed when I use fun2.

Why is this happening? What is the underlying difference between the two functions?

asked Oct 20 '21 at 19:10

Behzad Danaei

The first one simply assigns a new list to the local variable `a`, the other *mutates* whatever object `a` is referring to. – juanpa.arrivillaga Oct 20 '21 at 19:11
In `fun1`, you create a new object `a` rather than in `fun2`, you modify the values in `a` – Corralien Oct 20 '21 at 19:12
1

`a = ...` is a simple assignment, and makes the local variable named `a` refer to a different object. `a[0] = ...` is shorthand for `a.__setitem__(0, ...)`, which mutates the list already bound to `a`. – chepner Oct 20 '21 at 19:12
2

I would recommend reading this: https://nedbatchelder.com/text/names.html – jjramsey Oct 20 '21 at 19:12
Also, good idea to read the accepted answer to this question: https://stackoverflow.com/questions/986006/how-do-i-pass-a-variable-by-reference – juanpa.arrivillaga Oct 20 '21 at 19:15
@BehzadDanaei also, see the linked duplicate – juanpa.arrivillaga Oct 20 '21 at 19:16

Learning python: Why these two function yield different results?

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