如何将 say_whee 函数转换为 lambda？

Question

我想将 say_whee 变成 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()

我试过这样做，如下所示。不过好像没有调用函数。

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.

为清楚起见，@装饰器只是应用包装函数的语法糖。

这个

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

与

相同

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

与

相同

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

如何将 say_whee 函数转换为 lambda？

How do I turn the say_whee function into a lambda?

python

lambda

decorator