“is” operator not working as intended

Question

Just have a look at this code:

import re

ti = \"abcd\"
tq = \"abcdef\"
check_abcd = re.compile(\'^abcd\')
print id(check_abcd.search(ti))
print id(check_abc         


        

Answer 1

Martjin has given you the correct answer, but for further clarification here is an annotated version of what's happening in your code:

# this line creates returns a value from .search(), 
# prints the id, then DISCARDS the value as you have not 
# created a reference using a variable.
print id(check_abcd.search(ti)) 

# this line creates a new, distinct returned value from .search()
# COINCIDENTALLY reusing the memory address and id of the last one.
print id(check_abcd.search(tq))

# Same, a NEW value having (by coincidence) the same id
print check_abcd.search(ti)

# Same again
print check_abcd.search(tq)

# Here, Python is creating two distinct return values.  
# The first object cannot be released and the id reused
# because both values must be held until the conditional statement has 
# been completely evaluated.  Therefore, while the FIRST value will
# probably reuse the same id, the second one will have a different id.
if check_abcd.search(ti) is check_abcd.search(tq):
    print "Matching"
else:
    print "not matching"