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

Patricia Arquette
Release: 2024-11-01 19:11:30
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Why Does `(1 in [1, 0] == True)` Evaluate to False While `(1 in [1, 0]) == True` Evaluates to True?

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

When encountering the expression (1 in [1,0] == True), one might expect it to evaluate to True as in the case of (1 in [1,0]). However, it surprisingly evaluates to False. To understand this curious behavior, it's crucial to delve into Python's comparison operator chaining.

Python's operator chaining mechanism interprets the expression as:

<code class="python">(1 in [1, 0]) and ([1, 0] == True)</code>
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This expression has cascaded logical operators, with the "and" operator taking precedence. Consequently, the evaluation proceeds as follows:

  • (1 in [1, 0]) evaluates to True.
  • ([1, 0] == True) evaluates to False.

Since the "and" operator requires both operands to be True for a True result, and one of the operands is False, the entire expression evaluates to False.

This chaining behavior also applies to other chained comparison operators, such as:

  • a < b < c becomes (a < b) and (b < c)
  • a == b == c becomes (a == b) and (b == c)

Understanding this operator chaining mechanism is essential to correctly interpret and modify expressions involving cascaded comparison operators in Python.

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