**python : passing list as argument and updating in definition

Shell Programming and Scripting
1

In the below python code..
Could anyone please let me know why the name(variable) is getting modified if I update the kargs variable in the definition,

def f( kargs):
    kargs.extend([10])
    print ("In function :",kargs)

name = [1,2,3]
f(name)
print("Outside function :",name)

Output

>>> 
In function : [1, 2, 3, 10]
Outside function : [1, 2, 3, 10]
>>>
2

Python doesn't make a copy of the array when passed into a function, it passes a reference to the same array. More fundamental types would get passed by value instead.

1 Like
3

Because:

kargs.extend([10])

Which then changes name , which is displayed twice.

AFAIK: Python does preserve the value/content of variables among functions of the same file.
So changing it once, will effect all output of name , unless you set it back to its original value, before extending it, using a 2nd variable.

hth

4

But logically it is going to be wrong if I am not wrong.Is there a way to correct this one.

Logically speaking ... name and kargs are two different variables referring the same namespace.In our case .. since I am extending the local variable inside the function it should be local to the function itself.

But why it is effected the actual value.

Thanks much for reverting me for my question.

5

Because it actually holds an object ID or something like it.

6

In Python the assignment operator `=' never copies data. Whatever is at the left of the `=' is a binding label to a value, not a variable per se. This is significant, since moving labels do not produce another variable.

I recommend you watch Ned Batchelder - Facts and Myths about Python names and values - PyCon 2015 if you are interested on it.
At 15:50 you'll get presentation related to your query, but I would watch it from the beginning since the concept of binding and rebinding labels builds from understanding mutable and immutable values.

Hope it helps.