python getattr built-in method executes default arguments

问题

I dont know if this is something an expected behavior of getattr built_in method. getattr executes the default(3rd) argument as well even if the actual argument(2nd) satisfies the condition. Example:

def func():
    print('In Function')

class A:
    def __init__(self):
        self.param = 12

a = A()

When I run getattr(a, 'param', func()) it gives this result:

In Function
12

Note the In Function which I don't want.

But it works perfectly fine when I execute getattr(a, 'param1', func()) i.e output

In Function

But I only want result as 12 if satisfied the condition. Please let me know why getattr has such behaviour and can we stop them of doing it (that is not to execute 3rd arg if has 2nd argument), would be appreciated if share the alternate way of doing it in Pythonic way. One thing that comes in mind first is to check if param1 exist using hasattr and then do the needful.

回答1:

Before getattr gets executed, all the passed parameters have to be evaluated. func() is one of those parameters and an attempt to evaluate it will execute the print statement. Whether the attribute will be found or not, func() must be evaluated apriori.

This isn't peculiar to getattr, it's how functions and their parameters work in Python.

Consider the following:

>>> def does_nothing(any_arg): pass
...
>>> def f(): print("I'll get printed")
...
>>>
>>> does_nothing(f())
I'll get printed

Function does_nothing actually does nothing with the passed parameter. But the parameter has to be evaluated before the function call can go through.

The print statement however will not affect the outcome of getattr; sort of a side effect. In the event the attribute is not found the return value of the function is used.

来源：https://stackoverflow.com/questions/38589823/python-getattr-built-in-method-executes-default-arguments

标签

python

python-2.7

getattr