Local variables in Python nested functions

Answer

The nested function looks up variables from the parent scope when executed, not when defined.

The function body is compiled, and the 'free' variables (not defined in the function itself by assignment), are verified, then bound as closure cells to the function, with the code using an index to reference each cell. pet_function thus has one free variable (cage) which is then referenced via a closure cell, index 0. The closure itself points to the local variable cage in the get_petters function.

When you actually call the function, that closure is then used to look at the value of cage in the surrounding scope at the time you call the function. Here lies the problem. By the time you call your functions, the get_petters function is already done computing it's results. The cage local variable at some point during that execution was assigned each of the 'cow', 'dog', and 'cat' strings, but at the end of the function, cage contains that last value 'cat'. Thus, when you call each of the dynamically returned functions, you get the value 'cat' printed.

The work-around is to not rely on closures. You can use a partial function instead, create a new function scope, or bind the variable as a default value for a keyword parameter.

  • Partial function example, using functools.partial():

    from functools import partial
    
    def pet_function(cage=None):print"Mary pets the "+ cage.animal +"."yield(animal, partial(gotimes, partial(pet_function, cage=cage)))
  • Creating a new scope example:

    def scoped_cage(cage=None):def pet_function():print"Mary pets the "+ cage.animal +"."return pet_function
    
    yield(animal, partial(gotimes, scoped_cage(cage)))
  • Binding the variable as a default value for a keyword parameter:

    def pet_function(cage=cage):print"Mary pets the "+ cage.animal +"."yield(animal, partial(gotimes, pet_function))

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