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Home/ Questions/Q 56137
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Asked: May 10, 20262026-05-10T17:29:14+00:00 2026-05-10T17:29:14+00:00

Where X is any programming language (C#, Javascript, Lisp, Perl, Ruby, Scheme, etc) which

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Where X is any programming language (C#, Javascript, Lisp, Perl, Ruby, Scheme, etc) which supports some flavour of closures.

Some limitations are mentioned in the Closures in Python (compared to Ruby’s closures), but the article is old and many limitations do not exist in modern Python any more.

Seeing a code example for a concrete limitation would be great.

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  1. 2026-05-10T17:29:15+00:00Added an answer on May 10, 2026 at 5:29 pm

    The most important limitation, currently, is that you cannot assign to an outer-scope variable. In other words, closures are read-only:

    >>> def outer(x):  ...     def inner_reads(): ...         # Will return outer's 'x'. ...         return x ...     def inner_writes(y): ...         # Will assign to a local 'x', not the outer 'x' ...         x = y ...     def inner_error(y): ...         # Will produce an error: 'x' is local because of the assignment, ...         # but we use it before it is assigned to. ...         tmp = x ...         x = y ...         return tmp ...     return inner_reads, inner_writes, inner_error ...  >>> inner_reads, inner_writes, inner_error = outer(5) >>> inner_reads() 5 >>> inner_writes(10) >>> inner_reads() 5 >>> inner_error(10) Traceback (most recent call last):   File '<stdin>', line 1, in <module>   File '<stdin>', line 11, in inner_error UnboundLocalError: local variable 'x' referenced before assignment 

    A name that gets assigned to in a local scope (a function) is always local, unless declared otherwise. While there is the ‘global’ declaration to declare a variable global even when it is assigned to, there is no such declaration for enclosed variables — yet. In Python 3.0, there is (will be) the ‘nonlocal’ declaration that does just that.

    You can work around this limitation in the mean time by using a mutable container type:

    >>> def outer(x): ...     x = [x] ...     def inner_reads(): ...         # Will return outer's x's first (and only) element. ...         return x[0] ...     def inner_writes(y): ...         # Will look up outer's x, then mutate it.       ...         x[0] = y ...     def inner_error(y): ...         # Will now work, because 'x' is not assigned to, just referenced. ...         tmp = x[0] ...         x[0] = y ...         return tmp ...     return inner_reads, inner_writes, inner_error ...  >>> inner_reads, inner_writes, inner_error = outer(5) >>> inner_reads() 5 >>> inner_writes(10) >>> inner_reads() 10 >>> inner_error(15) 10 >>> inner_reads() 15 
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