Parameterized functions of no arguments?

[Rotwang]

On 11/02/2011 08:19, Steven D'Aprano wrote:
>
> [...]
>
> Re-writing them as normal function defs may help. I'm going to do them in
> reverse order:
>
> # lambda: f(k)
> def func():
>      return f(k)
>
> When you call func() with no arguments, the body is executed and f(k) is
> returned. Where do f and k come from? At runtime, Python searches the
> local namespace of func(), and doesn't find either f or k. It then
> searches the non-local namespace, that is, the function or method that
> surrounds func (or your lambda), if any. In your case, there is no such
> nested function, so finally it searches the global namespace, and finds
> both f and k. But by the time the function is called, the for-loop which
> sets k has reached the end, and k always has the same value.
>
>
>
> # (lambda x: lambda: f(x))(k)
> def func(x):
>      def inner():
>          return f(x)
>      return inner
>
>
> When you call func(k), it creates a nested function. That nested function
> includes a "closure", which is a copy of the namespace of func at the
> time it is called. This closure includes a variable "k".
>
> That inner function is returned and saved as the callback function. When
> you call that callback function, it executes inner(), and f(k) is
> returned. Where do f and k come from? As normal, Python searches the
> local namespace, and doesn't find them, but then it finds the variable k
> in the closure, and *not* the global k that the earlier example would
> find.
>
> This example may help:
>
> store = []
> for k in range(3):
>      fa = lambda: "k has the value %d" % k
>      fb = (lambda x: lambda: "x has the value %d" % x)(k)
>      print("fa inside the loop", fa())
>      print("fb inside the loop", fb())
>      store.append(fa)
>      store.append(fb)
>
> for func in store:
>      print("func outside the loop", func())
>
> del k
>
> store[1]()  # one of the closures
> store[0]()  # one of the non-closures
>
>
>
>
> Hope this helps.

Indeed it does. Thanks, and likewise to everyone else who replied.