Python- Adding odd numbers in a range including the limits?

Question

I need to write a program where the user inputs 2 numbers and then it gives the sum of all odd numbers in that range plus the 2 numbers that the user entered. I've searched around a lot but haven't found anything that includes the limits. So far I have:

x=int(input('Enter first number: '))
y=int(input('Enter second number: '))

def SumOdds(x,y):
  count=0
  for i in range(x,y):
     if (int(i%2==1)):
        count=count+i

  print(count)
SumOdds(x,y)

This gives the sum of the odds, but doesn't include the limits. For example, say I put in 10 and 20. This gives me 75, but it needs to add on the 10 and 20 to make it 105. I'm sure this is a simple fix, but I'm very new to Python so any help would be appreciated. Thanks!

Answer 1

So all you're missing is the addition of your two numbers when you finish your for loop. Try this:

x=int(input('Enter first number: '))
y=int(input('Enter second number: '))

def SumOdds(x,y+1):
  count= x + y #notice instead of 0, it's the sum now!
  for i in range(x,y):
     if(i == x or i == y):
         pass
     elif (int(i%2==1)):
        count=count+i

  print(count)
SumOdds(x,y)

Edit: as per your comment, you won't want to add your limits twice if they're odd. The y+1 ensures you're capturing the whole range, and the check for i == x or i == y skips those values in the range, since we've already added them at the start.

Answer 2

Just check the bounds separately.

x=int(input('Enter first number: '))
y=int(input('Enter second number: '))

def SumOdds(x,y):
  count=0
  for i in range(x,y):
    if (int(i%2==1)):
      count=count+i
  if(x%2==0):
    count= count+x
  if(y%2==0):
    count= count+7
  print(count)
SumOdds(x,y)

The base loop should include any odd limits, so you only have to add the limits if they are odd.