装饰师：为什么我得到不同的结果？如果我传递一个带有参数的函数，@ 符号是强制性的吗？

Question

def smart_divide(func):
def inner(a,b):
print("I am going to divide",a,"and",b)
if b == 0:
print("Whoops! cannot divide")
return
return func(a,b)
return inner
@smart_divide
def divide(a,b):
return a/b
print(divide(10,2))   # prints 5, correct

代码段2：

def smart_divide(func):
def inner(a,b):
print("I am going to divide",a,"and",b)
if b == 0:
print("Whoops! cannot divide")
return
return func(a,b)
return inner
def divide(a,b):
return a/b
print(smart_divide(divide(10,2))) # why is this printing function address ?

Answer 1

第二个片段是错误的。

@smart_divide
def divide(a,b):
...
divide(10,2)

相当于

smart_divide(divide)(10, 2)
# Not to smart_divide(divide(10,2))

看到区别了吗？smart_divide需要一个函数。你给它提供了一个浮子。如果你尝试再次调用它，你会得到预期的错误'float' object is not callable:

smart_divide(divide(10, 2))(10, 2)
TypeError: 'float' object is not callable