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#2
October 30th, 2003, 12:52 PM
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Code:
#!/usr/bin/env python
import rand
def getroll():
die = rand.rand() % 6 + 1
return die
die1 = getroll()
die2 = getroll()
print "Die 1: ", die1, " Die 2: ", die2, " Total: ", die1 + die2
__________________
Up the Irons What Would Jimi Do? Smash amps. Burn guitar. Take the groupies home. "Death Before Dishonour, my Friends!!" - Bruce D ickinson, Iron Maiden Aug 20, 2005 @ OzzFest Down with Sharon Osbourne Puzzle of the Month solved by sizeablegrin, etienne141 and L7Sqr, superior C/C++ programmers of the month
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#3
October 30th, 2003, 01:31 PM
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You could've used
Code:
die = random.randint(1,6) 
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#4
October 30th, 2003, 03:37 PM
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Infact Scorpi your dice role program wont work at all on newer versions of Python since the rand module no longer exists in the standard library.. What version of Python are you running ATM?
Mark.
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#5
October 30th, 2003, 03:50 PM
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Just for those who are interested, Python 1.5.2 was the last version of Python to support the rand module apparently. It seems to have been depreciated since even before version 2!
Quote:
It is still in Pythons doc's, in the "Undocumented Modules" part which simply says..Quote:
Take care, Mark. Last edited by netytan : October 30th, 2003 at 03:58 PM.
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#6
October 30th, 2003, 04:27 PM
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Actually, it happens that the box I tested the code snippet on is running Python 1.5.2 (on RH 7.1
 ). It's the only box I have lying around the office that has python on it. My boxes at home have later versions
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#7
October 30th, 2003, 05:28 PM
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Ah that explains it then
, i'd tell you to update but your one of the redhat guys and we don't wana break anything ..Well if were all in agreement you can do the dice thing like this mike: Code:
#!/usr/bin/env python
import random
#role the dice
die = random.randint(1, 6)
#deside what to do depending on th value of the die. i.e 1, 2, 3, 4, 5, 6
if die == 1:
print 'do soemthing when random is', random
if die == 2:
print 'do soemthing else when it is', random
else:
print 'do this if random is', random
well fix it then ,have fun, Mark.
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