Python "in" and "==" confusion

Question

print('a' in 'aa')
print('a' in 'aa' == True)
print(('a' in 'aa') == True)
print('a' in ('aa' == True))

The output is

True
False
True
Traceback (most recent call last):
  File "main.py", line 6, in <module>
    print('a' in ('aa' == True))
TypeError: argument of type 'bool' is not iterable

If line 2 is neither line 3 nor line 4, then what is it? How does it get False?

Answer 1

According to Expressions

print('a' in 'aa' == True)

is evaluated as

'a' in 'aa' and 'aa' == True

which is False.

See

print("a" in "aa" and "aa" == True)

==> False

The rest is trivial - it helps to keep operator precedence in mind to figure them out.

---

Similar ones:

• Multiple comparison operators in single statement
• Why does the expression 0 < 0 == 0 return False in Python?

with different statements. I flagged for dupe but the UI is wonky - I answered non the less to explain why yours exactly printed what it did.

Answer 2

Case 1 : it's simple the answers is True.

print('a' in 'aa')

Case 2 : This operation is evaluated as 'a' in 'aa' and 'aa' == True, so obviously it will return false.

print('a' in 'aa' == True)

Case 3: Now because we have () enclosing ('a' in 'aa') and the precedence of () is highest among all so first 'a' in 'aa' is evaluated as True and then True == True

print(('a' in 'aa') == True)

Case 4 : Same as above because of precedence of (), its evaluated as 'aa' == True, which will result in error as it tries to apply in on a non iterable that is bool value.

print('a' in ('aa' == True))