python的type()函数

如何解决python的type()函数

下面是我试过的代码

 class try1 :
    a = 0

 Box = type(try1())
 print(Box)

这是输出

<class '__main__.try1'>

现在我稍微改变了我的代码

class try1:
   a = 0
    
Box = type(try1())
print((type(Box)))

这是输出然后

<class 'type'>

现在我知道 type() 函数用于返回给定参数的类类型。任何人都可以帮助我理解在已经存储了另一个类的类型（即 try1）的变量（即 Box ）上应用 type() 函数的原因是什么？

解决方法

首先，type 不是一个函数，而是一个类。见the docs

也许这段代码让它更清晰一点：

class try1 :
    a = 0


instance = try1()
box = type(instance)
print(box)
print((type(box)))

# the type of the instance object is it's class
assert box is try1
# but classes themselves are objects too
assert isinstance(try1,object)
# and because box is try1
assert type(box) is type(try1)
# to be more specific,the class of a class is type
assert isinstance(try1,type)
assert type(try1) is type
# and therefore
assert repr(type(try1)) == "<class 'type'>"


# we could have created the try1 class by using the type class:
try1 = type('try1',(),{'a': 0})
# and end up with exactly the same thing
assert try1().a == 0
assert repr(type(try1())) == "<class '__main__.try1'>"

# well almost... we just created a new class and rehung the try1 to point to it
# box still points to the old try1
# so they may look the same
assert repr(try1) == repr(box)  # "<class '__main__.try1'>"
# they are different classes
assert not (try1 is box)

Python 中的所有变量都只是指向某事物的名称。 class try1: 只是一种语法糖，用于创建一个 __name__ 为 try1 的类型，并让命名空间中的名称 try1 指向它。