Re^3: Modulus zero error

Cool

The answer is perl has built a hash table in your while statement but when it comes to if it see's $p.. which is a new variable accordint to perl.. So there comes the problem. This is why it states the error in a wrong place

Note:
Use Strict and warnings

Update:
OOps sorry i mis understood the problem. Ok The problem lies in this line

         if (($n < $p[$i]) && ($x > $p[-1])) # error here
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just remove the previous code and use this u will not see the warning.

if (($n < $p[$i]))
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This is becasue the lt operator should work by finding the modulus and then determine the solution.(I am not sure with that but for this error i hope this could be the reason. i will update on you again after 15 minutes

Sorry i was not able to find it out....

Thanks
Sasi Kumar

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Re^4: Modulus zero error by Anonymous Monk on Dec 24, 2004 at 06:29 UTC
Eek... I don't understand, got lost at "built a hash table in your while statement" :( (I'm new to perl and programming in general, don't really know all that much about it yet...) I just want to know why `if (($n < $p[$i]) & ($x > 2))` gives me errors and `if (($n < $p[$i]) & ($x))` doesn't I tried use strict, warnings, didn't tell me anything different	[reply] [d/l] [select]
Re^5: Modulus zero error by maa (Pilgrim) on Dec 24, 2004 at 07:52 UTC
Can I ask... are you trying to do a Logical short-circuit AND (&&) in your if statement (to make sure both components are "true") or are you intentionaly performing a binary AND (&) on the data?	[reply]
Re^6: Modulus zero error by Anonymous Monk on Dec 24, 2004 at 08:31 UTC
Nope, that was just me being ignorant... I changed it to &&, but it still behaves the same way	[reply]
Re^7: Modulus zero error by maa (Pilgrim) on Dec 24, 2004 at 08:46 UTC
Re^4: Modulus zero error by ysth (Canon) on Dec 24, 2004 at 05:50 UTC
I didn't understand that at all; could you try to say it again?	[reply]