Python type() function returns a callable

Let's run through some
examples:

/>

>>>
            type(int)
            'type'>

So, since
type(int) returns a type, it makes
sense that

>>>
            type(int)(12)
            'int'>

since

>>>
            type(12)
            'int'>

More
importantly:

>>>
            (type(int)(12) == type(12)) and (type(int)(12) is
            type(12))
True

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Now, if you instead
do:

>>>
            type(int())
            'int'>

which
is also expected
since

>>> (int() == 0)
            and (int() is
            0)
True

and

>>>
            (type(int()) = type(0)) and (type(int()) is
            type(0))
True

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So, putting things together:

• int
  is an object of type
  type


• int()
  is an (integer) object of type
  int

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Another
example:

>>>
            type(str())
            'str'>

which means
that

>>>
            (type(str())() == '') and (type(str())() is
            '')
True

therefore,
it behaves just like a string
object:

>>>
            type(str())().join(['Hello', ', World!'])
'Hello,
            World!'

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I have the feeling that I might have made this seem
much more complicated than it actually is... it
isn't!

/>

type() returns the class of
an object.
So:

• type(12)
  is just an ugly way of writing
  int


• type(12)(12)
  is just an ugly way of writing
  int(12)

So,
"Yes!", type() returns a callable. But it's much better to
think about it as (from href="https://docs.python.org/3/library/functions.html#type" rel="nofollow
noreferrer">official docs )

readability="7">

class
type(object)

With one argument, return the type of an object. The return value is a type
object and generally the same object as returned by
object.__class__.