what is wrong with the if statement in the code below?

Question

can someone explain what's wrong with the below code?

     def factorial(num):
        if num == 0 or 1:
           return 1
        else:
           result = num * factorial(num-1)
           return result

#print(factorial(30))
#output: 1

& on the contrary when the if statement is slightly modified as below. The code seems to do what its meant to do. I would appreciate if someone could comment on why the code below provides desired output while the code above doesn't

     def factorial(num):
        if num == 1 or 0:
           return 1
        else:
           result = num * factorial(num-1)
           return result
 #print(factorial(30))
 #265252859812191058636308480000000

This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first. — PROTOCOL, Sep 25 '18 at 11:01

samwalton · Accepted Answer · 2018-09-25T10:56:49.393

0

num == 0 or 1 isn't doing what you think it is.

This could instead be num == 0 or num == 1 or num in [0,1].

edited Sep 25 '18 at 10:56

answered Sep 25 '18 at 10:55

samwalton

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1

Of course it's valid syntax, it just doesn't do what they expect – roganjosh Sep 25 '18 at 10:56
1

Ah, you're right, poor wording. Updated. – samwalton Sep 25 '18 at 10:57
That's useful feedback, thanks :) – samwalton Sep 25 '18 at 13:17
2

@samwalton what you said is what I already know. I'm not used to wording my questions in a way search would suggest to me what I'm looking for. It's particularly hard when one is short on time and has a ton to do. – Sahas Sep 25 '18 at 14:59

what is wrong with the if statement in the code below?

1 Answers1