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in reply to Just Another Pricey Holiday

Your code is a bit flawed. You are not guaranteed to have $some_play=1 after you buy expensive things. Allow me to alter your code a bit:
#!/usr/bin/perl # Just Another Pricey Holiday # # My beloved's pricey holiday # Pushing for expensive things always # But I bought those gifts # In double shifts # So tonight I'll guarantee some play... # # -higle my $beloveds_pricey_holiday = "Valentine's Day"; until ( $some_play ) { push @expensive_things, buy(qw{those gifts}); $some_play = 1 if &whim_of_significant_other; } sub buy() { push @shopping_cart, shift; push @shopping_cart, shift; } sub whim_of_significant_other { if (rand > 0.9) { return 1; } else { return 0; } }
Of course, your mileage may vary. ;)

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Re: Re: Just Another Pricey Holiday
by ehdonhon (Curate) on Feb 15, 2002 at 14:57 UTC

    I think your code is closer, but perhaps still a bit optimistic. Because the assignment of $some_play is in an until loop, it will continuously loop through and eventually get set (assuming that your randomizer has been seeded sufficiently so that it is possible for the whim to change).

    However, I've found that often buy() can somehow get stuck in an infinite loop until everything crashes. Once you are stuck in the loop, control is never passed back so that $some_play can even be considered. This can crash in two ways. Either you'll get a heap overflow from the recursion, or the credit card will fail.

    sub buy () { { $::not_a_good_night ||= rand; if ( $credit_card_balance > $credit_card_limit ) { die ( "Dog house tonight" ); } push @shopping_cart, shift(), shift(); buy(qw( more things )) if ( $::not_a_good_night > 0.3 ); }