如何将say_whee函数转换为lambda函数？

Question

我想把say_whee变成一个lambda函数。

def my_decorator(func):
    def wrapper():
        print("Something is happening before the function is called.")
        func()
        print("Something is happening after the function is called.")
    return wrapper

@my_decorator
def say_whee():
    print("Whee!")

if __name__ == '__main__':
    say_whee()

我试着这样做，如下所示。但是，它似乎没有调用该函数。

def my_decorator(func):
    def wrapper():
        print("Something is happening before the function is called.")
        func()
        print("Something is happening after the function is called.")
    return wrapper

@my_decorator
(my_decorator(lambda x: print (x))) ("Whee")

Answer 1

Lambda只是编写函数定义的快捷方式。

Lambda版本

say_whee = lambda : print ("Whee")

应用装饰器

say_whee = my_decorator(lambda : print ("Whee"))

完整代码

def my_decorator(func):
    def wrapper():
        print("Something is happening before the function is called.")
        func()
        print("Something is happening after the function is called.")
    return wrapper

say_whee = my_decorator(lambda : print ("Whee"))

if __name__ == '__main__':
    say_whee()

输出

Something is happening before the function is called.
Whee
Something is happening after the function is called.

为了清楚起见，@装饰器只是应用包装器函数的语法糖。

这

@my_decorator
def say_whee():
    print("Whee")

等同于

def say_whee():
    print("Whee")
    
say_whee = my_decorator(say_whee)

等同于

say_whee = my_decorator(lambda : print ("Whee"))