Simulating pointer in python.
Mutable and immutable value in memory
For immutable value like integer, consider the code below:
a = 10
b = a
b = 20
After that, a will equal to 10 and b will equal to 20. The pictures demonstrate the memory behind the code:
We can find the name of the value is actually a pointer, but we can’t use them in immutable value, because we can’t change the value pointed.
For mutable value like list, consider the code below:
a = [1,2,3,4]
b = a
b.append(5)
After that, both a and b will be [1,2,3,4,5], the memory usage is like:
By this property in mutable value, we can simulate pointer like c++.
Using mutable value simulate pointer
Take linked list by python as an example. This is the second question in LeetCode, Add two numbers.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
head = ListNode()
p1 = head
p2 = (l1,l2)
d = 0
while True:
if p2[0] and p2[1]:
value, s = ((p2[0].val + p2[1].val + d)%10),((p2[0].val + p2[1].val + d)//10)
p1.next = ListNode(value)
p1 = p1.next
p2 = (p2[0].next,p2[1].next)
d = s
elif p2[0]==None and p2[1]!=None:
value, s = ((p2[1].val + d)%10,(p2[1].val + d)//10)
p1.next = ListNode(value)
p1 = p1.next
p2 = (p2[0],p2[1].next)
d = s
elif p2[0]!=None and p2[1]==None:
value, s = ((p2[0].val + d)%10,(p2[0].val + d)//10)
p1.next = ListNode(value)
p1 = p1.next
p2 = (p2[0].next,p2[1])
d = s
elif p2[0]==None and p2[1]==None and d!=0:
p1.next = ListNode(d)
d=0
elif p2[0]==None and p2[1]==None and d==0:
break
return head.next
In this example, p1 and p2 are pointers.
