Why does the `is` operator behave differently with non-cached integers inside and outside functions in Python?

The is operator behaves unexpectedly with non-cached integers

The is operator in Python is used to compare the identity of two objects, meaning it checks if they refer to the same object in memory. However, when dealing with non-cached integers, the is operator can behave in an unexpected way.

If you evaluate the is operator on two integers within a function, it will return True, indicating that they refer to the same object. However, if you evaluate the same is operator outside the function, it will return False, indicating that they refer to different objects.

This is because of the way Python handles integer caching. For integers within the range [-5, 256], Python caches the integer objects and reuses them for subsequent references. This means that when you assign an integer within this range to multiple variables within a function, they all refer to the same cached object, and hence the is operator returns True.

However, when you assign an integer outside a function, it creates a new integer object and does not use the cached object. Therefore, when you compare two integers outside a function, they refer to different objects, and the is operator returns False.

To avoid this unexpected behavior, it is recommended to use the equality operator (==) to compare the values of integers, rather than the is operator.

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