如何在python中定义等效的私有方法

Question

i am new to python from java.

In java we would have something like

public void func1()
{
    func2();
}

private void func2(){}

However, in python i would like the equivalent

def func1(self):
    self.func2("haha")
    pass

def func2(str):
    pass

It throws me an error of takes exactly 1 argument(2 given)

I have checked for solutions such as using

def func1(self):
    self.func2("haha")
    pass

@classmethod 
def func2(str):
    pass

but it does not work

Taking out the self in func2 makes the global name func2 not defined.

How do i exactly resolve this situation.

Answer 1

Usually you do something like this:

class Foo(object):
    def func1(self):
        self._func2("haha")

    def _func2(self, arg):
        """This method is 'private' by convention because of the leading underscore in the name."""
        print arg

f = Foo()  # make an instance of the class.
f.func1()  # call it's public method.

Note that python has no true privacy. If a user wants to call your method, they can. It's just a fact of life that is described by the mantra "We're all consenting adults here" . However, if they call your method that is prefixed with an underscore, they deserve any breakages that they cause.

Also note that there are two levels of privacy:

def _private_method(self, ...):  # 'Normal'
    ...

def __private_with_name_mangling(self, ...):  # This name gets mangled to avoid collisions with base classes.
    ...

More info can be found in the tutorial .

Answer 2

Another possibility is that you want func2 to be private to (only exist in scope of) func1. If so, you can do this:

def func1(self, args):
    def func2(args):
        # do stuff
        pass
    return func2(args)

Answer 3

Try this:

class Foo:
  def method(self):
    self.static_method("haha")

  @classmethod 
  def static_method(clazz, str):
     print(str)

>>> Foo().method()
haha
>>> Foo.static_method('hi')
hi