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Posted on 2000-04-18
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Last Modified: 2010-03-05
1 Write a program in perl that prompts and restricts the user to any year between 1900
  and 2099, and for that year prints the sign of the Chinese Zodiac. Your
  solution can only use "if-elsif-else" type statements to implement the
  selections.  There are twelve signs in the Chinese Zodiac and they are,
  in order:

  Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig

  In addition, the Cat is used instead of the Rabbit in Vietnamese culture. If
  the year maps to Rabbit, print both.

  This sequence, starting with Rat, begins in the year 1900.  As a test, this
  year, 2000 is Year of the Dragon.
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Question by:myoreorex
2 Comments
 
LVL 16

Accepted Solution

by:
maneshr earned 200 total points
ID: 2727869
try this...........

#!/usr/local/bin/perl

@yrs=("Rat","Ox","Tiger","Rabbit","Dragon","Snake","Horse","Goat","Monkey","Roo
ster","Dog","Pig");

$sign_change=11;

print "Enter a year between 1900 & 2099 ";
$year=<STDIN>;
chomp($year); ## Remove the \n from the users input
$year=~ s/\s+//g; ## Remove any white spaces from the user's input.
if ($year<1900 || $year >2099){
  print "Year can ONLY be between 1900 and 2099\n";
  exit;
}

$sign=$year-1900;

while($sign>$sign_change){
  $sign-=scalar @yrs;
}

print "$year is the year of $yrs[$sign] ";

if ($yrs[$sign]=~ /^Rabbit$/){
  print "Cat";
}
0
 

Author Comment

by:myoreorex
ID: 2727905
Ok...I left out the chomp...thats where I went wrong.  Thanks!


Michelle
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