Python有像Perl 5.10的“状态”变量吗？

Question

In Perl 5.10, I can say:

sub foo () {
  state $x = 1;
  say $x++;
}

foo();
foo();
foo();

...and it will print out:

1
2
3

Does Python have something like this?

Answer 1

A class may be a better fit here (and is usually a better fit for anything involving "state"):

class Stateful(object):

    def __init__(self):
        self.state_var = 0

    def __call__(self):
        self.state_var = self.state_var + 1
        print self.state_var

foo = Stateful()
foo()
foo()

Answer 2

The closest parallel is probably to attach values to the function itself.

def foo():
    foo.bar = foo.bar + 1

foo.bar = 0

foo()
foo()
foo()

print foo.bar # prints 3

Answer 3

Python has generators which do something similar:

What does the "yield" keyword do in Python?

Answer 4

Not sure if this is what you're looking for, but python has generator functions that don't return a value per se, but a generator object that generates a new value everytime

def gen():
   x = 10
   while True:
      yield x
      x += 1

usage:

>>> a = gen()
>>> a.next()
10
>>> a.next()
11
>>> a.next()
12
>>> a.next()
13
>>> 

look here for more explanation on yield :
What does the "yield" keyword do in Python?

Answer 5

Here's one way to implement a closure in python:

def outer():
    a = [4]
    def inner():
        print a[0]
        a[0] = a[0] + 1
    return inner

fn = outer()
fn() # => 4
fn() # => 5
fn() # => 6

I borrowed this example verbatim from a python mailing list post .

Answer 6

Yes, though you have to declare your global variable first before it is encountered in foo :

x = 0

def foo():
    global x
    x += 1
    print x

foo()
foo()
foo()

EDIT: In response to the comment, it's true that python has no static variables scoped within a function. Note that x in this example is only exposed as global to the rest of the module. For example, say the code above is in test.py . Now suppose you write the following module:

from test import foo
x = 100
foo()
foo()

The output will be only 1 and 2 , not 101 and 102 .

Answer 7

You could also use something like

def static_num2():
    k = 0
    while True:
        k += 1
        yield k

static = static_num2().next

for i in range(0,10) :
    print static()

to avoid a global var. Lifted from this link about the same question.

Answer 8

Not that I'm recommending this, but just for fun:

def foo(var=[1]):
    print var[0]
    var[0] += 1

This works because of the way mutable default arguments work in Python.

Answer 9

The preferable way is to use class or generator ( yield ).

For the sake of completeness here's a variant w/ closure in Python 3.x:

>>> def make_foo():
...     x = 1
...     def foo():
...         nonlocal x
...         print(x)
...         x += 1
...     return foo
...
>>> foo = make_foo()
>>> foo()
1
>>> foo()
2
>>> foo()
3

Answer 10

>>> def foo():
    x = 1
    while True:
        yield x
        x += 1


>>> z = iter(foo())
>>> next(z)
1
>>> next(z)
2
>>> next(z)
3

Answer 11

Here's another dirty cheap way to do it, it's a variation on Tryiptich's answer, but using decorators

def static_var( name, value ):
    def dec( function ):
        setattr( function, name, value )
        return function
    return dec


@static_var( 'counter', 0 )
def counting_function():
    counting_function.counter = counting_function.counter + 1
    print counting_function.counter



"""
>>> counting_function()
1
>>> counting_function()
2
>>> counting_function()
3
>>> counting_function()
4
>>> counting_function()
5
>>> 
"""