Why Does (1 in [1, 0] == True) Evaluate to False in Python?

Python Precedence and Chaining: Why the Unexpected Result in (1 in [1,0] == True)?

In Python, the expression (1 in [1,0] == True) evaluates to False, which can seem counterintuitive. To understand why, it's essential to know about operator precedence and chaining in Python.

Operator precedence determines the order in which different operators in an expression are evaluated. In this case, the in operator has higher precedence than the comparison operator ==. So, the expression is parsed as (1 in [1,0]) == True, where (1 in [1,0]) is evaluated first.

When the interpreter evaluates (1 in [1,0]), it returns True because 1 is in the list [1,0]. However, the expression does not end there.

Python employs chaining for comparison operators. Chaining allows for multiple comparisons to be performed in a single expression. In this case, the expression (1 in [1,0]) == True can be translated into:

(1 in [1,0]) and ([1,0] == True)

After evaluating (1 in [1,0]) to True, the interpreter proceeds to evaluate ([1,0] == True). This comparison evaluates to False because [1,0] is a list and True is a boolean value.

Therefore, the overall expression evaluates to False because both conditions in the chained comparison are not met. It's not simply a matter of precedence, but rather the combination of precedence and chaining that leads to this result.

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