[Tutor] Help with if-elif-else structure

[Prasad, Ramit]

>if roll[0] > 15:
>   .........
>elif roll[0] <=15:  <--- this is redundant. already checked for > 15 so if here is always <= 15
>    if roll[1] >= 2:
>
>you can change with:
>
>elif roll[1] >= 2:

That is not true. The first one looks at index [0], while the second one is index[1]. 

>>if roll[0] > 15:
>>   if roll[1] >= 2:
>>       print("Success")
>>   elif roll[2] >= 2:
>>       print("Critical failure!")

This will print "Success" if the first roll is greater than 15 AND second roll is 2 or greater. 

It will print "Critical failure!" if the first roll is greater than 15 AND the second roll is less than 2 AND the third roll is 2 or greater.

Ramit


Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology
712 Main Street | Houston, TX 77002
work phone: 713 - 216 - 5423


From: tutor-bounces+ramit.prasad=jpmorgan.com at python.org [mailto:tutor-bounces+ramit.prasad=jpmorgan.com at python.org] On Behalf Of TheIrda
Sent: Friday, August 26, 2011 8:55 AM
To: Christopher King
Cc: python mail list
Subject: Re: [Tutor] Help with if-elif-else structure

Hello,

I'm quite new on python, so don't hit too hard if I'm wrong ;)


A question on the logic... Does this means

if roll[0] > 15:
   if roll[1] >= 2:
       print("Success")
   elif roll[2] >= 2:
       print("Critical failure!")


that rolling multiple 1's get the priority on rolling multiple 6's? I mean result > 15 and 2 1's is always a critical success even if multiple 6 are rolled?

also

if roll[0] > 15:
   .........
elif roll[0] <=15:  <--- this is redundant. already checked for > 15 so if here is always <= 15
    if roll[1] >= 2:

you can change with:

elif roll[1] >= 2:



And... which is the RPG name ? :p
Cheers
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