Python: Compute and print sum of two given integers

Python Basic - 1: Exercise-41 with Solution

Write a Python program to compute and print sum of two given integers (more than or equal to zero). If given integers or the sum have more than 80 digits, print "overflow".

Sample Solution:

Python Code:

print("Input first integer:")
x = int(input())
print("Input second integer:")
y = int(input())
if x >= 10 ** 80 or y >= 10 ** 80 or x + y >= 10 ** 80:
    print("Sum of the two integers: ",x + y)

Sample Output:

Input first integer:
Input second integer:
Sum of the two integers:  47


Flowchart: Python - Compute and print sum of two given integers

Python Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous: Write a Python program to check if a point (x,y) is in a triangle or not. There is a triangle formed by three points.
Next: Write a Python program that accepts six numbers as input and sorts them in descending order.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

Python: Tips of the Day

Iterating over dictionaries using 'for' loops:

I am a bit puzzled by the following code: d = {'x': 1, 'y': 2, 'z': 3} for key in d: print key, 'corresponds to', d[key] What I don't understand is the key portion. How does Python recognize ...

key is just a variable name.

for key in d:

For Python 3.x:

for key, value in d.items():

For Python 2.x:

for key, value in d.iteritems():

To test for yourself, change the word key to poop.

In Python 3.x, iteritems() was replaced with simply items(), which returns a set-like view backed by the dict, like iteritems() but even better. This is also available in 2.7 as viewitems().

The operation items() will work for both 2 and 3, but in 2 it will return a list of the dictionary's (key, value) pairs, which will not reflect changes to the dict that happen after the items() call. If you want the 2.x behavior in 3.x, you can call list(d.items()).

Ref: https://bit.ly/37dm0Qo