使用变量定义多个函数docstring

Question

我有几个函数(a、b和c)，我希望它们使用相同的docstring。因此，我的计划是保存行，只编写一次docstring，并将其保存到变量DOCSTRING中。然后我把它放在函数声明下。我在PEP 257上没有发现任何能解决我的问题的东西.

DOCSTRING = '''
This is a docstring
for functions:
a,
b,
c'''

def a(x, y):
    DOCSTRING
    # do stuff with x and y

def b(x, y):
    DOCSTRING
    # do other stuffs with x and y

def c(x, y):
    DOCSTRING
    # do some more stuffs with x and y

help(a), help(b), help(c)

实际上我想这可能是work...but我错了，我明白了：

Help on function a in module __main__:

a(x, y)

Help on function b in module __main__:

b(x, y)

Help on function c in module __main__:

c(x, y)

这一点用处都没有。

我进行了第二次尝试，将函数的特殊__doc__属性更改为docstring DOCSTRING。

DOCSTRING = '''
This is a docstring
for functions:
a,
b,
c'''

def a(x, y):
    a.__doc__ = DOCSTRING # also tried just __doc__ = DOCSTRING, both didn't work
    # do stuff with x and y

def b(x, y):
    b.__doc__ = DOCSTRING # also tried just __doc__ = DOCSTRING, both didn't work
    # do other stuffs with x and y

def c(x, y):
    c.__doc__ = DOCSTRING # also tried just __doc__ = DOCSTRING, both didn't work
    # do some more stuffs with x and y

help(a), help(b), help(c)

这两种方法的输出结果与之前的尝试相同.

目前起作用的是好的复制粘贴方法：

def a(x, y):
    '''
    This is a docstring
    for functions:
    a,
    b,
    c'''

    # do stuff with x and y

def b(x, y):
    '''
    This is a docstring
    for functions:
    a,
    b,
    c'''

    # do other stuffs with x and y

def c(x, y):
    '''
    This is a docstring
    for functions:
    a,
    b,
    c'''

    # do some more stuffs with x and y

help(a), help(b), help(c)

当然，它产生了我想要的输出：

Help on function a in module __main__:

a(x, y)
    This is a docstring
    for functions:
    a,
    b,
    c

Help on function b in module __main__:

b(x, y)
    This is a docstring
    for functions:
    a,
    b,
    c

Help on function c in module __main__:

c(x, y)
    This is a docstring
    for functions:
    a,
    b,
    c

如你所见，但这样做会迫使我不得不浪费多行来写同样的东西.

所以，现在，我的问题是，如果我将docstring复制到每个函数，，而不需要将它复制到每个函数，那么如何才能得到相同的结果呢？

Answer 1

您的问题是，尝试在函数体中设置docstring将无法工作，因为除非函数实际被调用，否则这些行永远不会被计算。你需要的是类似(或相当于)的东西：

def c(x, y):
    # code

c.__doc__ = DOCSTRING
help(c)

你会想要用一个装饰师。

def setdoc(func):                                                                                                                                                                                                            
    func.__doc__ = DOCSTRING                                                                                                                                                                                                 
    return func

@setdoc                                                                                                                                                                                                                
def c(x, y):                                                                                                                                                                                                                 
    print("hi")                                                                                                                                                                                                              

help(c) #Output docstring