8 Réponses
+ 1
In python, all objects are transfered by reference. However, some objects are immuable, and their modification implies the creation of a modified copy of the object. Any user-defined data-type is by default considered as mutable.
You can use the ctypes library, that contains a real pointer datatype but... I'm not sure it is a good way.
If you want to simulate pointers, on basic types, you could do wrappers around native objects :
class Wrapper:
def __init__(self, value):
self.value = value
But there is no real way to have pointers in Python.
+ 1
Lists act as a pointer. Use list for call by reference and normal variable passing for call by value.
Sure lists do take a lot of memory but it'll help you.
e.g.
def func(a):
a[0] += 1
a = [5]
func(a)
print(a[0])
Link to code for your reference: https://code.sololearn.com/cc1A5NDY3Svw/#py
+ 1
Milind Yadav it is not a matter of 'normal variable' or 'list'. It is a matter of mutability, as I explained previously. And you're actually saying the same thing as I said : creating a wrapper object around another object.
+ 1
Yep.. that's true.... unfortunately there is no other way....this was just a short code.... that's why I mentioned this
0
We are using the OOPS concept to overcome this problem
0
J Santhosh Kumar But OOP is also available in c++, I don't think that can be used here. Please read the question again.
0
Yeah but In python there is no pointers. They using only OOPS concept.
0
Yes, I'd loved to see pointers too... But at the end, I never really needed them in Python.