how to return last condition if other 2 not met?

[MRAB]

On 2018-05-24 00:57, asa32sd23 at gmail.com wrote:
> i want to check/return for 3 conditions, it loops shortest str and finds diff in other
> 1. if difference is immediate before end of range, return index, exit
> 2. if string length is same and index loop is done, return 'identical'
> 3. if neither of above is found. it means the short loop ended and every letter was same so next letter of longer str is the diff, just return idex+1
> but my last print statement always print. not sure how to end this
> 
> [code]
> str1= "kitti cat"
> str2= 'kitti catt'
> lenStr1= len(str1)
> lenStr2= len(str2)
> 
> #find shortest str and loop range with this one
> if lenStr1 >= lenStr2:
>      str= str2
> else:
>      str= str1
>          
> # loop each character of shortest string, compare to same index of longer string
> # if any difference, exit and return index of difference
> for idx in range(len(str)):
>      a= str1[idx]
>      b= str2[idx]
>      if a != b:      #immeditely exit, since non-match found
>          print(idx)
>          break
>      else:
>         if len(str1) == len(str2) and idx == len(str1)-1: #if no difference print 'identical'
>              print("identical")
>              break
> 
> print(idx+1)
> [/code]
> 
The 'for' loop (and also the 'while' loop) can have an 'else' clause, 
which is run if it didn't break out of the loop:

for idx in range(len(str)):
     # body of loop
     ...
else:
     # didn't break out of the loop
     print(idx+1)