Python static variable generator function

Question

I am using the following function to generate a static variable in Python:

 #Static variable generator function
def static_num(self):
    k = 0
    while True:
        k += 1
        yield k

When I am calling this function from main code:

 regression_iteration = self.static_num()
 print " Completed test number %s  %s \n\n" % (regression_iteration, testname)

I get this output:

  "Completed test number <generator object static_num at 0x027BE260>  be_sink_ncq"

Why I am not getting an incremented integer? Where is my static variable generator going wrong?

Edit:

I am calling the function static_num in the following manner now:

regression_iteration = self.static_num().next()

But it returns only '1' since the value of 'k' is being initialized to zero every time the function is called. Therefore, I do not get the required output 1,2,3,4 .... on every call of the function

What are you trying to do with this generator? And how does your concept of a static number work? — jedwards
– jedwards, Commented Mar 17, 2015 at 6:25
"Yield is a keyword that is used like return, except the function will return a generator." You are printing the whole object not just the values. See this post. — Marco
– Marco, Commented Mar 17, 2015 at 6:29
When you say static variable, do you mean a variable that is shared between all instances of a class? If so, how does your generator help accomplish this? — jedwards
– jedwards, Commented Mar 17, 2015 at 6:32
Similar to the concept of static variable in C, I need a variable whose value will remain unchanged when a function is called repeatedly. At the end of the function, I will increase the value of the variable, so that in the next function call it gets an incremented value and does not get reset to zero like an automatic variable — user3565150
– user3565150, Commented Mar 17, 2015 at 6:39
Why not just not modify the value inside the function. I may be misunderstanding, but it seems like you're doing significantly more work than necessary. — jedwards
– jedwards, Commented Mar 17, 2015 at 6:49

jedwards · Accepted Answer · 2015-03-17 06:58:00Z

4

Its hard to say whether you need to use this approach -- I strongly doubt it, but instead of a generator you could abuse using mutable types as default initializers:

def counter(init=[0]):
    init[0] += 1
    return init[0]

x = counter()
print(x)  # 1
print(x)  # 1
print(x)  # 1
x = counter()
print(x)  # 2
print(x)  # 2
print(x)  # 2
# ... etc

The return value of counter increases by one on each call, starting at 1.

answered Mar 17, 2015 at 6:58

jedwards

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3 Comments

user3565150 Over a year ago

jedwards, this worked perfectly !! Thanks a lot for your support.

jedwards Over a year ago

@user3565150 No problem, and sorry for the confusion in comments. And thank you for pushing me over 10k rep :)

user3565150 Over a year ago

Its fine, you have helped me a lot before, I can take this little pain :)

amow · Accepted Answer · 2015-03-17 06:26:34Z

0

The yield keyword convert your function return value from int to an int generator, so you have to use "for i in self.static_num()" to access the int.If you want to access the generator one by one. Use next function.

answered Mar 17, 2015 at 6:26

amow

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Comments

Matt · Accepted Answer · 2015-03-25 02:42:24Z

0

Assuming that by static you meant static to the class, the best way to do this is to use a class variable and a classmethod:

class Static_Number_Class(object):
    _static_number = 0
    @classmethod
    def static_number(cls):
        cls._static_number += 1
        return cls._static_number

If this is just a stand alone function you could use a closure instead:

def Counter():
    n = 0
    def inner():
        n += 1
        return n
    return inner

count = Counter()
x = count()
print(x) # 1
x = count()
print(x) # 2

edited Mar 25, 2015 at 2:42

answered Mar 25, 2015 at 0:20

Matt

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