How to write a simple callback function?

Question

Python 2.7.10

I wrote the following code to test a simple callback function.

def callback(a, b):
    print('Sum = {0}'.format(a+b))

def main(callback=None):
    print('Add any two digits.')
    if callback != None:
        callback

main(callback(1, 2))

I receive this when I execute it:

Sum = 3
Add any two digits.

Why Add any two digits is after Sum = 3? I guess it is because the callback function executes first. How to execute the callback function after all other code in main() executed?

Answer 1

In this code

if callback != None:
    callback

callback on its own doesn't do anything; it accepts parameters - def callback(a, b):

The fact that you did callback(1, 2) first will call that function, thereby printing Sum = 3, and then main() gets called with the result of the callback function, which is printing the second line

Since callback returns no explicit value, it is returned as None.

Thus, your code is equivalent to

callback(1, 2)
main()

---

Solution

You could try not calling the function at first and just passing its handle.

def callback(n):
    print("Sum = {}".format(n))

def main(a, b, _callback = None):
    print("adding {} + {}".format(a, b))
    if _callback:
        _callback(a+b)

main(1, 2, callback)

Answer 2

Here's what you wanted to do :

def callback(a, b):
    print('Sum = {0}'.format(a+b))

def main(a,b,f=None):
    print('Add any two digits.')
    if f is not None:
        f(a,b)

main(1, 2, callback)