I'm trying to make the code comparable under the same conditions as possible, but no matter how many times I try, the list is rewritten in-place. Why?I would appreciate it if the answer was anything other than "You should use tuples instead of lists."
Also, since this is my first question post, I think there are some rude points, so it would be even more helpful if you could point that out.Thank you for your cooperation.
>>>lst,a=[0],0
>>def test1(lst):
try:
_=lst
_[0] = 1
except:
pass
return_
>>def test2(x):
try:
b = a
b = 1
except:
pass
return b
>> test1(lst)
[1]
>>>lst
[1] # I want it to remain 0 here.
>> test2(a)
1
>>a
0
In the code in question, id(test1(lst)) and id(lst) are identical and refer to the same.
The code for the question should be as follows.
def test1(lst):
try:
_=lst.copy()
_[0] = 1
except:
pass
return_
There is an explanation in Python's official FAQ, so you should read it first:
Simply supplement, when you create an object like a list, you create a single entity that stores the contents of the element, and when you argument or substitute a function for the object itself, you do not create a new entity, but a reference to point to it.
Therefore, when passing a list to a function argument, to prevent the function to which you pass from modifying the original list, instead of passing lst to the argument, you can pass lst[:] or lst.copy() to create a new list.Alternatively, you can create and pass a tuple that is an array that cannot be modified, such as tuple(lst) (in this case, the variable that receives the tuple will fail if it tries to change its contents).
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