On 2021-10-03 10:39:38 +1100, Chris Angelico wrote: > # Oh, and it doesn't actually assign anything. > >>> def f(n): > ... return (lambda: (n := 1)), (lambda: n) > ... > >>> g, h = f(5) > >>> h() > 5 > >>> g() > 1 > >>> h() > 5
I thought about that one a bit more and I think that the lambda is creating a new scope, just like a regular function would, so it is assigning to its own private «n», not the «n» in «f(n)». https://docs.python.org/3/reference/expressions.html#grammar-token-lambda-expr doesn't mention scope explicitely, but | The unnamed object behaves like a function object defined with: | | def <lambda>(parameters): | return expression implies to me that it should create a new scope because a nested function defined with «def» would do that, too. hp -- _ | Peter J. Holzer | Story must make more sense than reality. |_|_) | | | | | h...@hjp.at | -- Charles Stross, "Creative writing __/ | http://www.hjp.at/ | challenge!"
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