cls()函数在类方法中做了什么？

Question

今天我正在查看另一个代码,看到了这个:

class A(B): 
    # Omitted bulk of irrelevant code in the class

    def __init__(self, uid=None):
        self.uid = str(uid)

    @classmethod
    def get(cls, uid):
        o = cls(uid)
        # Also Omitted lots of code here

这个cls()功能在这做什么？

如果我让其他类继承这个A类,调用它C,在调用这个get方法时,这会o使用Cclass作为调用者cls()吗？

Answer 1

cls是构造函数,它将构造类A并调用该__init__(self, uid=None)函数.

如果你使用它(使用C),cls将保持'C',(而不是A),请参阅AKX答案.

Answer 2

对于classmethods,第一个参数是通过其调用类方法而不是通常self用于instancemethods的类(除非另有说明,否则隐式使用类中的所有方法).

这是一个例子 - 为了练习,我添加了一个检查cls参数身份的异常.

class Base(object):
    @classmethod
    def acquire(cls, param):
        if cls is Base:
            raise Exception("Must be called via subclass :(")
        return "this is the result of `acquire`ing a %r with %r" % (cls, param)

class Something(Base):
    pass

class AnotherThing(Base):
    pass

print Something.acquire("example")
print AnotherThing.acquire("another example")
print Base.acquire("this will crash")

this is the result of `acquire`ing a <class '__main__.Something'> with 'example'
this is the result of `acquire`ing a <class '__main__.AnotherThing'> with 'another example'
Traceback (most recent call last):
  File "classmethod.py", line 16, in <module>
    print Base.acquire("this will crash")
  File "classmethod.py", line 5, in acquire
    raise Exception("Must be called via subclass :(")
Exception: Must be called via subclass :(

Answer 3

这是一个班级工厂.

基本上它与调用相同:

o = A(uid)

cls在def get(...): 是 A.