is — Expert Playground

Identity operator; tests whether two variables reference the same object

Python Playground

import dis

def check_is(x):
    return x is None

def check_is_not(x):
    return x is not None

def check_eq(x):
    return x == None

print("x is None:")
dis.dis(check_is)
print("\nx is not None:")
dis.dis(check_is_not)
print("\nx == None:")
dis.dis(check_eq)

# 'is' compiles to IS_OP (fast pointer comparison)
# '==' compiles to COMPARE_OP (calls __eq__)
# IS_OP has two forms: IS_OP 0 (is) and IS_OP 1 (is not)

Output

Click "Run" to execute your code

'is' compiles to IS_OP which is a simple pointer comparison (O(1)). '==' compiles to COMPARE_OP which calls __eq__ and may be arbitrarily slow. 'is not' is a single opcode, not 'not' applied to 'is'.

Challenge

Try modifying the code above to explore different behaviors. Can you extend the example to handle a new use case?

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