#                          CSC 15 Lab 9

#                        Due Next Thursday


######## Basic mechanics of object oriented programming.  ########

## Reread the notes and examples that I gave you last week on OOP to freshen
#  your memory.  Here's another example below:

# Sample class construction.  Early in the semester we wrote a program that
# added minutes and seconds.  The values for minutes and seconds were
# kept in different variables.  We are now going to rewrite the program
# in the OOP style.  Such an organization will allow the program to be usable
# in a variety of situations (i.e., as parts of larger programs).

## I must first *design* my objects by being clear as to:
#   1. What are the pieces of data that the object will consist of.  In other
#      words: what variables will belong to each object.
#      Answer: minutes, seconds
#   2. What methods (functions) do I wish to call on each object.
#      Answer: *add*: add two minutes. *increment*: increment time by one sec.
#      *reset*: sets time to 0:0.  *tostring*: represent time in string format

# Now I can write my class of time objects:

class mstime:	  # minutes-seconds time obects

      def __init__(self,m,s):	    # constructor initializes variables.
	  self.min = m		    # initialize minutes
	  self.sec = s		    # and seconds
      # constructor

      def reset(self):   # resets values to 0 min, 0 sec.
	  self.min = 0
	  self.sec = 0
      # reset

      def inc(self):	# increment time destructively by one second.
	  self.sec += 1 # destructive means the 'self' object is changed.
	  if (self.sec>=60):
	     self.min = self.min + self.sec / 60
	     self.sec = self.sec % 60
	  # if 
      # inc

      def add(self, B):  # non-destructively add self and B time objects
	  s = self.sec + B.sec
	  m = self.min + B.min
	  extramin = s / 60
	  s = s % 60
	  m = m + extramin
	  newtime = mstime(m,s)  # new object to be returned
	  return newtime 
      # add

      # The function above needs to be called in the form
      # myobject.add(yourobject).  For common operators like +, -, *, etc..
      # it's possible to redefine the operators for your class using the
      # keywords __add__, __sub__, __mul__ and __div__:
      def __add__(self, B):  # redefines + operator for mstime objects
	  s = self.sec + B.sec
	  m = self.min + B.min
	  extramin = s / 60
	  s = s % 60
	  m = m + extramin
	  newtime = mstime(m,s)  # new object to be returned
	  return newtime
      # add

      def tostring(self):   # convert to string for display representation
	  return str(self.min) + ":" + str(self.sec)
      # tostring
# end mstime

### usage of class mstime:
t1 = mstime(3,40)  # 3 min, 40 sec.
t2 = mstime(1,59)  # 1 min, 50 sec.
t2.inc()
t3 = t1.add(t2)
t4 = t3.add(t1)  
print t2.tostring()
print t3.tostring()
print t4.tostring()

t5 = mstime(3,50) + t3   # implicitly calls t5.__add__(t3)
print t5.tostring()


#############################################################################
## Part I:

## I'm a gas station owner and I've hired you to program for me.  Each
#  gas station sells only one type of gas: regular unleaded.  
#  Each station will have an initial supply of 100 gallons of gas.
#  Each station also has a manager, who will be fired if sales aren't good.
#  I want a computer to keep track of the sales and supply situation at each
#  gas station.  I want to be able to run the following code:


# Create station objects with manager's name and initial price per gallon:
st1 = gasstation("john makeabuck",3.39)   
st2 = gasstation("jane gouger",3.79)   

st1.sellgas(10.5)   # station1 just sold 10.5 gallons of gas
st2.sellgas(20)     # station2 just sold 20 gallons of gas
st1.sellgas(3000)   # should print not enough gas, no sale.
st1.sellgas(11)
print st1.totalsales()  # prints the dollar amount of total sales. (3.39*21.5)
print st1.currentsupply() # prints gallons of gas left in supply.
st1.resupply()      # reset gas supply to 100 gallons
st2.sale()          # decrease gas price by 10%
st2.gouge()         # double price during natural disasters
st2.replacemanager("jimmy manageron")  # fire old manager, hire jimmy.

if st1.moresales(st2): print "station1 has better sales than station2"

## Now write the class to make this work.  Think first: what variables 
# will each class contain?


###### Part II: Dates. #####################################################

# For the second class, you need to implement objects similar to the mstime
# object.  You want a class of objects called 'date'.  Each date object will
# have a year, a month, and a day.  Here's the code you'd like to write:

date1 = date(4,15,2008)    # create date object for april 15th, 2008
date2 = date(12,31,2004)    # december 31st, 2004

print date1.tostring()  # prints "4/15/2008"
print date2.leapyear()  # prints True, since 2004 % 4 == 0
date1.nextday()         # advance date (destructively) to 4/16/2008
date2.nextday()         # advance date to 1/1/2005
if date2.before(date1): print "date2 comes before date1"  #### don't forget!

## Hint: the difficult method is nextday.  For this, you need to know
## the number of days in each month.  The following array may help you:

Days = [31,28,31,30,31,30,31,31,30,31,30,31]

# However, you also need to adjust the value of Days[1] (Feburary) if
# the current year is a leap year.  You can define this array inside your
# nextday function, or as a variable of your object (self.Days).  
# Alternatively, instead of using an array you can write another function
# (also inside your class) that returns the number of days given the month
# number.

# Hint for the 'before' method:  To check if one date comes before another,
# first compare the year.  If the years are the same, compare the month.  If
# the months are also the same, compare the day.  


#############################################

## (CHALLENGE) The formula for determining leap years is not as simple
# as dividing by 4.  Research the real formula and implement it.

## (MEGACHALLENGE): implement a method to compute the number of days until
#   the given date:

christmas = date(12,25,2008)
print date1.daysuntil(christmas), "days until Christmas"

# Hint: convert the date to one number: christmas is day 359 of 365.


## (GIGACHALLENGE) Write a method to return the day of the week
#  of a data object.  It should work as follows:

date1 = date(4,14,2008)
print date1.dayofweek()  # should print Monday

# Knowing that 4/14/2008 is a Monday should be enough for you to figure
# out the day-of-week of all the other dates.


## (TERACHALLENGE)  Friday the 13th is an unlucky day (and a very bad movie).
#  Write a function to find the NEXT friday the 13th after a given date:

date1.nextfri13()  # should return a date object representing the next
                   # friday the 13th after date1.