解释 probSecond.calls 如何等于零

Question

def MainCount(f):
    def progFirst(*args,**kwargs):
        progFirst.calls+=1
        return f(*args,**kwargs)
    progFirst.calls=0
    return progFirst
@MainCount
def progSecond(i):
    return i+1

@MainCount
def Count(i=0,j=1):
    return i*j+1
print(progSecond.calls)
for n in range(5):
    progSecond(n)
Count(j=0,i=1)
print(Count.calls)

输出：0 1

根据我的理解 MainCount(probSecond) 但我不明白 probSecond.calls 如何在 Count.calls 中也等于零

Answer 1

正如您在 MainCount 函数中看到的那样 probFirst.Calls 是函数的属性。当 MainCount(probSecond) 现在 probSecond.calls 也是 MainCount 函数的属性。

# A Python example to demonstrate that 
# decorators can be useful attach data 

# A decorator function to attach 
# data to func 
def attach_data(func): 
    func.data = 3
    return func 

@attach_data
def add (x, y): 
    return x + y 

# Driver code 

# This call is equivalent to attach_data() 
# with add() as parameter
print(add(2, 3)) 

print(add.data)

解释 probSecond.calls 如何等于零

Explain how probSecond.calls equal to zero

python

python-2.7

python-3.x

python-decorators