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Comment by Dylan16807

17 days ago

Even without laziness, you can get similar problems if f creates a closure or returns something that includes the parameter object.

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Dylan16807

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trealira  17 days ago

Yeah, this is a common source of confusion with closures in Python. Example on Stack Overflow:

https://stackoverflow.com/questions/233673/how-do-lexical-cl...

  • elbear  16 days ago

    Doesn't that example also show a kind of laziness?

    I say this because the second solution to that question offers the solution of using `i` as a default argument when defining the function. That forces its evaluation and fixes the problem.

    • trealira  16 days ago

      It's just a name shadowing.

      Copying the code they wrote:

        for i in xrange(3):
      def func(x, i=i): # the *value* of i is copied in func() environment
          return x * i
      flist.append(func)

      That could also be written "def func(x, foo=i): return x * foo". It's just copying i's value to another variable. In the next line, i's value is still 1, 2, or 3, so when the function is called during the next line of the body of the loop, the value held by i is bound to foo.

      It's not evaluating a thunk representing i, which is how lazy variables are evaluated in Haskell.

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