try / except python handling and int() function

Question

Fairly new to the Pythonic Way, but the below code doesn't work as expected, anyone know why? If I input 143.22 I get "not an int" response, rather than the expected collatz(143) e.g., int(143.22). BTW, this is one basic exercise in the Automate the Boring Stuff with Python beginners book (chapter 3 exercise on using functions). Thanks!

def collatz(myint) :
    print(myint)   
    if myint % 2 == 0 :
        myint = int(myint / 2)
    else:
        myint = int(myint *3 +1)
    if myint == 1 :
        print(myint)
        return myint
    else:
        collatz(myint)

try:
    x=input('Input a non-negative integer between 1 and infinity: ')
    val = int(x)
    if int(x) == abs(int(x)) :
        print("Yes input string is an Integer.")
        collatz(val)
    else:
        print('I did say NON negative!')
except ValueError:
    print("That's not a positive int!")

Indeed 143.22 isn't a positive int, it is a float. Do you want your program to handle both floats (by rounding?) and ints? — Kent Shikama
– Kent Shikama, Commented Dec 7, 2019 at 21:08
the error is caused from val = int(x). You can apply int(143.22), but you are not allowed to int('143.22'). Read more docs.python.org/3.8/library/functions.html#int — facehugger
– facehugger, Commented Dec 7, 2019 at 21:13
Also, you can replace int(x) == abs(int(x)) by val == abs(val). — sammy
– sammy, Commented Dec 7, 2019 at 21:17
It seems to me like your program is working fine. You are asking for a non-negative integer, and when giving it 143.22 it shouts "that's not an int". What seems to be the problem? — Tomerikoo
– Tomerikoo, Commented Dec 7, 2019 at 21:20
I believe @facehugger has it nailed; removing that (val=int(x)) line makes no difference in the functioning of the routine, so I consider this a learning exercise in type conversion/casting as well. — AlMo320
– AlMo320, Commented Dec 23, 2019 at 18:05

ashears · Accepted Answer · 2019-12-07 21:34:44Z

1

You are trying to run int('143.22')

For me, this yields:

Traceback (most recent call last):
File "main.py", line 1, in <module>
    int('143.22')
ValueError: invalid literal for int() with base 10: '143.22'

However, int(143.22) works fine.

A quick solution for your use case may be to cast it as a float.

float('143.22')

int(float('143.22'))

answered Dec 7, 2019 at 21:34

ashears

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1 Comment

AlMo320 Over a year ago

Thanks! Python appears to be somewhere a 'strongly typed' and 'loosely typed' language, my background with both Perl and C# seems to make the learning curve a bit longer, but ... worth the journey!!

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