id(object)
Returns unique identity ( integer or long integer ) of the python object. my_int=34
print(id(my_int)) # 10969856
Using list my_list=[1,2,3]
print(id(my_list)) #139787163392648
Below output will change each time we run the program but watch how the id remain same for different objects.
a=23
b=a
print(id(a)) #1173520149488
print(id(b)) #1173520149488
str1='plus2net'
str2='plus2net'
print(id(str1)) #1173522439088
print(id(str2)) #1173522439088
my_int=-5
print(id(my_int)) # 10968608
my_int=256
print(id(my_int)) # 10976960
my_int=257
my_tuple=(5,7,5,4,3,5)
print(id(my_tuple)) # 139787173397704
my_tuple +=(9,8)
print(id(my_tuple)) # 139787174601112
print(my_tuple) #(5, 7, 5, 4, 3, 5, 9, 8)
Using List
my_list=[5,7,5,4,3,5]
print(id(my_list)) # 139787163608584
my_list +=[9,8]
print(id(my_list)) # 139787163608584
print(my_list) # [5, 7, 5, 4, 3, 5, 9, 8]
Two non-overlapping lifetime objects may have the same identity or id() vlaue. my_list=[1,2,3]
str='welcome'
print("address of list : ", id(my_list))
print("address of string str : ",id(str))
# change the values
# modified the list object, not a new object , address is same.
my_list[1]=7
print("address of list : ", id(my_list))
# str[1]='p' # this will generate error
# new string object is created now,
# not modified the old one, address changed
str='Python'
print("address of string str : ", id(str))
Output
address of list : 140301581527624
address of string str : 140301582741432
address of list : 140301581527624
address of string str : 140302155689016
The id value of list is not changing, but it is changing for string. Author
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