Menu
All comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation. Thus "==" and "<" have the same priority, why would the first expression in the following evaluate to True, different from the 2nd expression?
>>> -1 < 0 == False
True
>>> (-1 < 0) == False
False
I would expect both be evaluated to False. Why is it not the case?
522
All comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation. Thus "==" and "<" have the same priority, why would the first expression in the following evaluate to True , different from the 2nd expression?
Summary. A comparison operator compares two values and returns a boolean value, either True or False . Python has six comparison operators: less than ( < ), less than or equal to ( <= ), greater than ( > ), greater than or equal to ( >= ), equal to ( == ), and not equal to ( != ).
The logical-AND operator ( && ) has higher precedence than the logical-OR operator ( || ), so q && r is grouped as an operand. Since the logical operators guarantee evaluation of operands from left to right, q && r is evaluated before s-- .
Answer: The correct order of precedence is given by PEMDAS which means Parenthesis (), Exponential **, Multiplication *, Division /, Addition +, Subtraction -.
Python has a really nice feature - chained comparison, like in math expressions, so
-1 < 0 == False
is actually a syntactic sugar for
-1 < 0 and 0 == False
under the hood.
145
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
