锁定函数的装饰函数？

Question

假设我有两个函数，它们都调用一个decorator函数，该函数以两个方法作为参数：

def decorator(connector, disconnector):
def decorator(function):
    def decorated_function(*args, **kwargs):
        print("connecting")
        connector(args[0])
        result = function(*args, **kwargs)
        print("disconnecting")
        # only want this to be called when function() is completely done
        disconnector(args[0])
        return result
    return decorated_function
return decorator

@decorator(connector, disconnector)
def func1():
   func2()
   print "func2 done"

@decorator(connector, disconnector)
def func2():
   print "hello world"

调用func1（）时，是否有方法防止在func1（）完全执行完毕之前调用装饰器中的隔离开关方法？在这种情况下，我不希望对装饰器的func2（）调用执行连接/断开连接

Answer 1

如果您希望在断开连接之前完全完成func1，请执行以下操作：

def connector(*args):
    print("connect done", args[0])
def disconnector(*args):
    print("disconnect done", args[0])


def decorator(connector, disconnector):
    def decorator(function):
        def decorated_function(*args, **kwargs):
            print("connecting")
            connector(args[0])
            result = function(*args, **kwargs)
            print("disconnecting")
            # only want this to be called when function() is completely done
            disconnector(args[0])
            return result
        return decorated_function
    return decorator

@decorator(connector, disconnector)
def func1(a):
   func2()
   print("func2 done")

def func2():
   print("hello world")

func1(10)

输出：

connecting
connect done 10
hello world
func2 done
disconnecting
disconnect done 10

注意：上面的代码是用python3编写的。在您的情况下，只需更新代码中的print语句（删除括号）

由于您正在将参数传递给connector(args[0])，请将参数传递给func1()。我传递了一个伪整数=10