function scope

[Steven D'Aprano]

On Mon, 02 Feb 2009 16:37:07 -0800, Mike Kent wrote:

> On Feb 2, 6:40 pm, Baris Demir <demirba... at gmail.com> wrote:
> 
>> def simpleCut(d=dict()):
>>        temp=d
>>        for i in temp.keys():
>>            if    (temp[i] == .......) :
>>               temp[i]=new_value
>> return temp
> 
> You have been bitten by the shared default parameter noobie trap:

No, he has tripped over the assignment-doesn't-make-copies feature. 
Ignore the default value, and consider passing in a dict:


def simpleCut(d=dict()):
    temp=d
    # ...

That doesn't make a temporary copy of d, it makes a new name that refers 
to the same dictionary as d. So if you do this:

li = {'x': 1}
f = simpleCut(li)

then

assert f is li

will pass, and he will be surprised and dismayed to discover that 
simpleCut() has side-effects.


Solution: make a copy of the input.


def simpleCut(d=dict()):
    temp = d.copy()
    for i in temp:  # don't need to call keys()
        if (temp[i] == ...):
            temp[i] = new_value
    return temp



-- 
Steven