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   veer66 (veer66@mstdn.in.th)'s status on Sunday, 30-Mar-2025 21:48:36 JST veer66

   From the statement below, can I infer that an if-block is a scope because it doesn't have a name, and therefore it is not directly accessible?

   > A scope is a textual region of a Python program where a namespace is directly accessible. “Directly accessible” here means that an unqualified reference to a name attempts to find the name in the namespace."

   https://docs.python.org/3/tutorial/classes.html#python-scopes-and-namespaces

   #python

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     sedje (sedje@fosstodon.org)'s status on Sunday, 30-Mar-2025 21:48:35 JST sedje

     in reply to

     @veer66 ?

     No. It has nothing to do with "naming" the if-block, but with resolving a variable name (the unqualified reference). The if-expression may or may not use variables, but that is irrelevant.

     An if-block does not create a separate scope.

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     veer66 (veer66@mstdn.in.th)'s status on Sunday, 30-Mar-2025 21:49:06 JST veer66

     in reply to

     • sedje

     @sedje

     > An if-block does not create a separate scope.

     Because a variable inside an if-block is directly accessible in the same namespace as it is outside the if-block?

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     sedje (sedje@fosstodon.org)'s status on Sunday, 30-Mar-2025 22:14:54 JST sedje

     in reply to

     @veer66 Correct. Generally only a function or class will open a new scope/namespace.

     But also take care with exception handler (argument)s and list comprehension and generator expressions.

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