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pythonoperator-precedence

Why does (1 == 2 != 3) evaluate to False in Python?

Why does (1 == 2 != 3) evaluate to False in Python, while both ((1 == 2) != 3) and (1 == (2 != 3)) evaluate to True?

What operator precedence is used here?

Solution

  • This is due to the operators' chaining phenomenon. The Python documentation explains it as:

    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).

    And if you look at the precedence of the == and != operators, you will notice that they have the same precedence and hence applicable to the chaining phenomenon.

    So basically what happens:

    >>>  1==2
=> False
>>> 2!=3
=> True

>>> (1==2) and (2!=3)
  # False and True
=> False