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如何从函数内部打印python函数的Docstring?

更新时间:2023-12-04 10:48:22

  def my_func():

print my_func .__ doc__

只要您不会更改绑定到名称 my_func 的对象。

  new_func_name = my_func 
my_func =无

new_func_name()
#不打印任何东西,因为my_func是None且None没有文档字符串

这是相当罕见的,但它们确实发生了。然而,如果你写这样的装饰器:



def passin(func):
def wrapper(* args,** kwargs):
返回func(func,* args,** kwargs)
返回包装

现在您可以执行此操作了:

  @passmein 
def my_func(me):
打印我.__ doc__

这将确保你的函数获得对自身的引用(类似于 self )作为它的第一个参数,所以它总是可以获得正确功能的文档字符串。如果使用方法,通常 self 成为第二个参数。


I want to print the docstring of a python function from inside the function itself. for eg.

def my_function(self):
  """Doc string for my function."""
  # print the Docstring here.

At the moment I am doing this directly after my_function has been defined.

print my_function.__doc__

But would rather let the function do this itself.

I have tried calling print self.__doc__ print self.my_function.__doc__ and print this.__doc__ inside my_function but this did not work.

def my_func():
    """Docstring goes here."""
    print my_func.__doc__

This will work as long as you don't change the object bound to the name my_func.

new_func_name = my_func
my_func = None

new_func_name()
# doesn't print anything because my_func is None and None has no docstring

Situations in which you'd do this are rather rare, but they do happen.

However, if you write a decorator like this:

def passmein(func):
    def wrapper(*args, **kwargs):
        return func(func, *args, **kwargs)
    return wrapper

Now you can do this:

@passmein
def my_func(me):
    print me.__doc__

And this will ensure that your function gets a reference to itself (similar to self) as its first argument, so it can always get the docstring of the right function. If used on a method, the usual self becomes the second argument.