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Asked: 2026-06-13T13:51:56+00:00

Possible Duplicate: Why does the expression 0 < 0 == 0 return False in

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Possible Duplicate:
Why does the expression 0 < 0 == 0 return False in Python?

The following output from the Python 2 REPL confuses me:

>>> 15>10==True
False
>>> 15>1==True
True
>>> 15>2==True
False
>>> 15>False
True

If 15>10==True is evaluated as (15>10)==True the expression would simplify to print True==True, which obviously evaluates to True. If 15>10==True is evaluated as 15>(10==True) the expression simplifies to 15>False which also evaluates to True. Both of these interpretations contradict the actual value of the expression (False).

I can understand 15>1==True evaluating to True since 1==True is true, but no interpretation of 15>10==True makes sense to me.

Summary: In Python 2, why does 15>10==True evaluate to False?

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1 Answer

  1. Editorial Team

    This is because chained comparison in Python. Namely, 15>10==True is actually evaluated as: 

    15 > 10 and 10 == True

    which is False.

    On the other hand, 15>1==True is the same as

    15 > 1 and 1 == True

    which evaluates to True.

    To quote from the docs:

    Unlike C, all comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation. Also unlike C, expressions like a < b < c have the interpretation that is conventional in mathematics:

    comparison    ::=  or_expr ( comp_operator or_expr )*
comp_operator ::=  "<" | ">" | "==" | ">=" | "<=" | "<>" | "!="
                   | "is" ["not"] | ["not"] "in"

    Comparisons yield boolean values: True or False.

    Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

    Formally, if a, b, c, …, y, z are expressions and op1, op2, …, opN are comparison operators, then a op1 b op2 c … y opN z is equivalent to a op1 b and b op2 c and … y opN z, except that each expression is evaluated at most once.

    Note that a op1 b op2 c doesn’t imply any kind of comparison between a and c, so that, e.g., x < y > z is perfectly legal (though perhaps not pretty).

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