Home > Article > Backend Development > Why Does Python\'s \"is\" Operator Behave Differently with Large Integers in Different Code Blocks?
In Python, the is operator checks object identity rather than equivalence. However, in certain scenarios, this behavior can be unexpected, as demonstrated below:
<code class="python">def func(): a = 1000 b = 1000 return a is b a = 1000 b = 1000 print(a is b, func()) # (False, True)</code>
The is comparison returns True within the function, but False outside it. Despite the different evaluation results, the integers a and b appear to have the same value.
To understand this behavior, the Python interpreter's interpretation of code blocks is crucial. According to the Python reference manual, each command executed interactively constitutes a separate code block. The following are considered blocks:
Within a single code block, Python caches certain objects, such as small integers (within the range [-5, 256]). When multiple variables refer to the same integer literal, they will all refer to the same cached integer object. However, if the literals are declared in different code blocks, distinct integer objects will be created.
To illustrate this, consider the code objects for each case:
As a result, a is b evaluates to True within the func block, where all integers are cached, and False outside the block, where separate objects are created.
Note the following caveats:
Understanding the concept of code blocks in Python is essential for interpreting unexpected behavior involving identity comparisons. While is is useful for checking the identity of immutable objects, it is recommended to use the equality operator (==) for comparing values, as it does not rely on caching.
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