Sum of square of first n natural numbers

Question

try:
num=int(input("Enter a number:"))

def sum(num):
result=0
if num < 0:
print(num, "is not a natural number!")
else:
for i in range(1,num+1):
result=result + (i*i)
return result

print("The sum of square of first", num, "natural number is:", sum(num))
except ValueError:
print("Invalid Input")

For the given code;
How can I not execute,
***print("The sum of square of first", num, "natural number is:", sum(num))***
this statement for input less than zero?

Putting this statement inside else block did not help!

Answer 1

Move the `sum` function out of the try ... except block, then do the validation to decide whether to call the function (input is natural number, > zero)

def sum(num):
// function body

try:
# input <num> variable
if <num> < 0:
# error message
else:
# call the sum function
# display square sum here
except ValueError:
# bad input ...