我怎样才能告诉PyCharm参数期望是什么类型？
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当涉及到构造函数，赋值和方法调用时，PyCharm IDE非常擅长分析我的源代码并弄清楚每个变量应该是什么类型。我很喜欢它，因为它给了我很好的代码完成和参数信息，并且如果我尝试访问一个不存在的属性，它会给我警告。
但是当涉及到参数时，它一无所知。代码完成下拉列表无法显示任何内容，因为它们不知道参数的类型。代码分析无法查找警告。
class Person:
def __init__(self, name, age):
self.name = name
self.age = age
peasant = Person(\"Dennis\", 37)
# PyCharm knows that the \"peasant\" variable is of type Person
peasant.dig_filth() # shows warning -- Person doesn\'t have a dig_filth method
class King:
def repress(self, peasant):
# PyCharm has no idea what type the \"peasant\" parameter should be
peasant.knock_over() # no warning even though knock_over doesn\'t exist
King().repress(peasant)
# Even if I call the method once with a Person instance, PyCharm doesn\'t
# consider that to mean that the \"peasant\" parameter should always be a Person
pygame.Surface
class King:
def repress(self, peasant):
\"\"\"
Exploit the workers by hanging on to outdated imperialist dogma which
perpetuates the economic and social differences in our society.
@type peasant: Person
@param peasant: Person to repress.
\"\"\"
peasant.knock_over() # Shows a warning. And there was much rejoicing.
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5 个回复
殿虫
朗排谎纷瘟
class King: def repress(self, peasant: Person) -> bool: peasant.knock_over() # Shows a warning. And there was much rejoicing. return peasant.badly_hurt() # Lets say, its not known from here that this method will always return a bool>>> King.repress.__annotations__ {\'peasant\': <class \'__main__.Person\'>, \'return\': <class \'bool\'>}苛肠倒俱
class King: def repress(self, peasant): \"\"\" Exploit the workers by hanging on to outdated imperialist dogma which perpetuates the economic and social differences in our society. @type peasant: Person @param peasant: Person to repress. \"\"\" peasant.knock_over() # Shows a warning. And there was much rejoicing.橙绥
class Replicant(object): pass class Hunter(object): def retire(self, replicant): \"\"\" Retire the rogue or non-functional replicant. :param Replicant replicant: the replicant to retire. \"\"\" replicant.knock_over() # Shows a warning.公藕
def my_function(an_int): assert isinstance(an_int, int) # Pycharm now knows that an_int is of type int pass