Python相当于golang的defer语句

Question

如何实现像deferpython中的语句一样工作的东西？

Defer将函数调用推送到堆栈.当包含defer语句的函数返回时,在defer语句首先位于内部的范围内,逐个弹出并执行deferred函数调用.延迟语句看起来像函数调用,但在弹出之前不会执行.

去举例说明它是如何工作的:

func main() {
    fmt.Println("counting")

    var a *int
    for i := 0; i < 10; i++ {
        a = &i
        defer fmt.Println(*a, i)
    }

    x := 42
    a = &x

    fmt.Println("done")
}

输出:

counting
done
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1
0 0

去用例的例子:

var m sync.Mutex
func someFunction() {
    m.Lock()
    defer m.Unlock()
    // Whatever you want, with as many return statements as you want, wherever.
    // Simply forget that you ever locked a mutex, or that you have to remember to release it again.
}

Answer 1

要模拟defer fmt.Println(*a, i)示例,您可以使用contextlib.ExitStack:

#!/usr/bin/env python3
from contextlib import ExitStack
from functools import partial

print("counting")
with ExitStack() as stack:
    for i in range(10):
        a = i
        stack.callback(partial(print, a, i))

    x = 42
    a = x
    print("done")

产量

counting
done
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1
0 0

可以轻松模拟互斥锁的情况:

def some_function(lock=Lock()):
    with lock:
        # whatever

Answer 2

Python的with语句与Go的延迟具有相似的目的.

Python中类似的代码是:

mutex = Lock()

def someFunction():
    with mutex:
        # Whatever you want, with as many return statements
        # as you want, wherever. Simply forget that you ever
        # locked a mutex, or that you have to remember to 
        # release it again.

Answer 3

我在那里做了一个(与2.x兼容):

@defers_collector
def func():
    f = open('file.txt', 'w')
    defer(lambda: f.close())

    defer(lambda : print("Defer called!"))

    def my_defer():
    recover()

    defer(lambda: my_defer())

    print("Ok )")
    panic("WTF?")

    print("Never printed (((")


func()
print("Recovered!")

来源defers_collector是:

# Go-style error handling

import inspect
import sys

def panic(x):
    raise Exception(x)

def defer(x):
    for f in inspect.stack():
    if '__defers__' in f[0].f_locals:
        f[0].f_locals['__defers__'].append(x)
        break

def recover():
    val = None
    for f in inspect.stack():
    loc = f[0].f_locals
    if f[3] == '__exit__' and '__suppress__' in loc:
        val = loc['exc_value']
        loc['__suppress__'].append(True)
        break
    return val

class DefersContainer(object):
    def __init__(self):
    # List for sustain refer in shallow clone
    self.defers = []

    def append(self, defer):
    self.defers.append(defer)

    def __enter__(self):
    pass

    def __exit__(self, exc_type, exc_value, traceback):
    __suppress__ = []
    for d in reversed(self.defers):
        try:
            d()
        except:
            __suppress__ = []
            exc_type, exc_value, traceback = sys.exc_info()
    return __suppress__


def defers_collector(func):
    def __wrap__(*args, **kwargs):
    __defers__ = DefersContainer()
    with __defers__:
        func(*args, **kwargs)
    return __wrap__

Answer 4

一个延迟部分的灵感来自于实施@DenisKolodin 答案是可以作为的一部分pygolang，2：

   wc = wcfs.join(zurl)    ?     wc = wcfs.join(zurl)
   defer(wc.close)         ?     try:
                           ?        ...
   ...                     ?        ...
   ...                     ?        ...
   ...                     ?     finally:
                           ?        wc.close()

Answer 5

对jfs 答案的补充ExitStack在装饰器的帮助下进一步推动了这个想法：

@with_exit_stack
def counting(n, stack):
    for i in range(n):
        stack.callback(print, i)


@with_exit_stack
def locking(lock, stack):
    stack.enter_context(lock)
    # whatever

with_exit_stack定义如下：

import functools
import contextlib

def with_exit_stack(func):
    @functools.wraps(func)
    def wrapper(*args, **kwargs):
        with contextlib.ExitStack() as stack:
            return func(*args, **kwargs, stack=stack)

    return wrapper