SOLUTION MANUAL FOR Data Abstraction & Problem Solving with C++: Walls and Mirrorsby Frank Carrano , Timothy Henry ISBN:978-0134463971 NEW COMPLETE GUIDE WITH RATIONALES 100% VERIFIED A+ GRADE ASSURED!!!!!NEW LATEST UPDATE!!!

Solutions to Selected Exercises
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(Version 7.0) aw




Data Abstraction & Problem Solving with C++
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Seventh Edition a w




Frank M. Carrano aw aw




University of Rhode Island
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Timothy M. Henry aw aw




New England Institute of Technology
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, 2


Solution Manual & Test Bank for Data Abstraction & Problem Solving with C++: Walls and Mirrors, 7th Edition by Frank M. Carrano
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Chapter 1 Data Abstraction: The Walls aw aw aw aw aw




1
const CENTS_PER_DOLLAR = 100;
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/** Computes the change remaining from purchasing an item costing dollarC
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ost dollars and centsCost cents with d dollars and c cents. Preconditi
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on: dollarCost, centsCost, d and c are all nonnegative integers and ce
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ntsCost and c are both less than CENTS_PER_DOLLAR. Postcondition: d a
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nd c contain the computed remainder values in dollars and cents respec
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tively. If input value d < dollarCost, the proper negative values for
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the amount owed in d dollars and/or c cents is returned. */
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void computeChange(int dollarCost, int centsCost, int& d, int& c);
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2a
const MONTHS_PER_YEAR = 12;
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const DAYS_PER_MONTH[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
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/** Increments the input Date values (month, day, year) by one day.
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Precondition: 1 <= month <= MONTHS_PER_YEAR, aw aw aw aw aw



1 <= day <= DAYS_PER_MONTH[month - 1], except aw aw aw aw aw aw aw



when month == 2, day == 29 and isLeapYear(year) is true. Postcondit aw aw aw aw aw aw aw aw aw aw aw



ion: The valid numeric values for the succeeding month, day,
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and year are returned. */ aw aw aw aw


void incrementDate(int& month, int& day, int& year);
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/** Determines if the input year is a leap year.
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Precondition: year > 0. aw aw aw



Postcondition: Returns true if year is a leap year; false otherwise. */ aw aw aw aw aw aw aw aw aw aw aw



bool isLeapYear(int year);
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3a
changeAppointmentPurpose(apptDate: Date, apptTime: Time, purpose: string): boolean aw aw aw aw aw aw



{
if (isAppointment(apptDate, apptTime)) cancelAppointment(apptDate
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, apptTime) aw




return makeAppointment(apptDate, apptTime, purpose)
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}




© 2017 Pearson Education, Inc., Hoboken, New Jersey 070
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30

, 3



3b
displayAllAppointments(apptDate: Date): void aw aw



{
time = START_OF_DAY aw aw



while (time < END_OF_DAY) aw aw aw



if (isAppointment(apptDate, time)) displa
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yAppointment(apptDate, time) aw



time = time + HALF_HOURaw aw aw aw



}
This implementation requires the definition of a new operation
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displayAppointment()
as well as definitions for the constants START_OF_DAY, END_OF_DAY and HALF_HOUR.
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4
// Assume that storeBag is defined and contains your purchased items Bag<std::string>
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fragileBag;
while (storeBag.contains("eggs"))
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{
storeBag.remove("eggs"); fragileBag.add("eggs" aw



);
} // end while
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while (storeBag.contains("bread"))
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{
storeBag.remove("bread"); fragileBag.add("brea aw



d");
} // end while
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// Transfer remaining items from storeBag to groceryBag; Bag<std::st
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ring> groceryBag; aw



v = storeBag.toVector();
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for (int i = 0; i < v.size(); i++) groceryBag.add
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(v.at(i));




© 2017 Pearson Education, Inc., Hoboken, New Jersey 070
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30

, 4



5
/** Removes and counts all occurrences, if any, of a given string from
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a given bag of strings.
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@param bag A given bag of strings. @pa
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ram givenString A string.
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@return The number of occurrences of givenString that occurred and
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were removed from the given bag. */
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int removeAndCount(ArrayBag<std::string>& bag, std::string givenString)
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{
int counter = 0; aw aw aw



while (bag.contains(givenString))
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{
counter++; bag.remove(givenString aw



);
} // end while
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return counter; aw



} // end removeAndCount
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6
/** Creates a new bag that combines the contents of this bag and a second ba
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g without affecting the contents of the original two bags.
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@param anotherBag The second bag.
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@return A bag that is the union of the two bags. */
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public BagInterface<ItemType> union(BagInterface<ItemType> anotherBag);
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7
/** Creates a new bag that contains those objects that occur in both this
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bag and a second bag without affecting the contents of the original two bags. @par
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am anotherBag The given bag.
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@return A bag that is the intersection of the two bags. */
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public BagInterface<ItemType> intersection(BagInterface<ItemType> anotherBag);
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8
/** Creates a new bag of objects that would be left in this bag after removing those ob
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jects that also occur in a second bag without the contents of the original two bag
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s.
@param anotherBag The given bag.
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@return A bag that is the difference of the two bags. */
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public BagInterface<T> difference(BagInterface<T> anotherBag);
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© 2017 Pearson Education, Inc., Hoboken, New Jersey 070
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