Problem 1 (10 points): Short Answer

Problem 1 (35 points): Short Answer

1. (4 points) Write the boolean expressions that represent the following conditions. Assume that the variables have been declared and initialized appropriately. Be careful to write out the expres= sions as you would type them in C++.

(= a) n is not between 5 and 15 inclusi= ve

Ans: !(5 <=3D n && n <= =3D 15)

(= b) Either x and y are both divi= sible by 3 or both divisible by 5 (remember, a % b gives the remainder when a is divided by= b)

(x % 3 =3D=3D 0 && y % 3 =3D= =3D 0) || (x % 5 =3D=3D 0 && y % 5 =3D=3D 0)

2. (6 points) The following expressions are intended to convert a distance measur= ed in light-years and stored in an int variable called = lyr to the equivalent distance in parsecs stored in an int variable called par. The expression should round to the nearest light-year. There are 3.25 (or ₌) light-years in a parsec. For each of the following expressi= ons, write the letter of the correct response in the blank provided:

a) = Does not produce the correct result

b) = Produces the correct result

i.&n= bsp; par =3D lyr / 3.25 + 0.5; = //Ans: b

ii.&= nbsp; par =3D (int) (lyr / 13 * 4 + 1 / 2); //Ans: a

iii.= par =3D (int) ((double)lyr * 4 / 13 + 0.5); //Ans: b=

(2 points) What happens when you open a write file with the parameter ios::binary, and why would = you want to do this? Ans: = It opens in binary mode which means that you will create a binary file. You do this to have a random access file.

= 4. (5 points) twoD is a 2-dimensional array of size n. Write a loop that calculates the p= roduct of the diagonal of int prod =3D 1;

for (int i =3D 0; i < n; = i++)

prod *=3D twoD[i][i];

cout << prod << = endl;

= 5. (3 points) Find the error in eac= h of the following segments. Assum= e the following declarations and statements:

int *zPtr; // zPtr will reference= array z

void *sPtr =3D 0;

int number;

int z[5] =3D {1 , 2, 3, 4, 5};=

sPtr =3D z;&nbs= p; zPtr =3D z;

= a) number =3D zPtr; // Use pointe= r to get first value of array

Ans: number =3D *zPtr;

= b) number =3D *zPtr[2]; // Assign array= element 2 to number

Ans: number =3D zPrt[2];

= c) number =3D *sPtr; // Assign th= e value pointed to be sPtr to number

Ans: number =3D *((int *) sPtr)

6. (5 points) Write a method containsCS that= returns the number of element= s that contain “cs” in an array of strings called sarr. The number of= elements in the array is given in the parameter size.

int containsCS(string [] sarr, int size)

int count =3D 0;<= /span>

for (int i =3D 0; i < siz= e; i++) {

if (sarr[i].find(“cs”)

= !=3D string::npos)

count++;

}

return count;

}

&n= bsp;

7. (10 points) Consider the following method, which is valid despite being poorly formatted:

public void foo(char ch)

{

if (ch !=3D ‘l’)=

if (ch >=3D ‘a̵= 7; && ch <=3D ‘z’)

if (ch !=3D ‘r’)=

cout << “dogR= 21; << endl;

else

cout << “catR= 21; << endl;

else

cout << “mouse&#= 8221; << endl;

}

<= /o:p>

(a) Rewrite the above method using proper indentation.

public void foo (char ch) {

if (ch !=3D ‘l’)=

if (ch >= ;=3D ‘a’ && ch <=3D ‘z’)

if (ch !=3D ‘r’)

= cout << “dog” << endl;

else

= cout << “cat” << endl;

else<= /o:p>

cout << “mouse” << endl;

}

(b)&= nbsp; Give a value of ch for which foo displays cat. = &nb= sp; //Ans: r

(c)&= nbsp; Give a value of ch for which foo displays dog. //Ans: a

(d)&= nbsp; Give a value of ch for which foo displays mouse. = //Ans: A

(e)&= nbsp; Give a value of ch for which foo does not display anything. //Ans: l=

Problem 2 (15 points): Complete the following program. It will first open a file called “physics.dat”. The first line of the file contains the initial velocity (v₀) and in= itial position (x₀) of an object, separated by a space. The object’s velocity (v) and position (x) were measured at regular intervals and recorded on subsequent lines as shown:

0.0004 1

0.00089 2

0.0019 2.5

…

Use the formula below to calculate the acceleration fo= r each measurement.

₌

Finally, the program writes a file “newPhys.dat&= #8221; which includes the acceleration as a third piece of data on each line as sh= own:

0.0004 1

0.00089 2 1.0535e-07

0.0019 2.5 3.28571e-07=

…

#include <fstream> // Necessary header file=

#include <= ;cmath>

<= /o:p>

using std::e= ndl;

<= /o:p>

int main() {=

<= /o:p>

// Declare variables

double<= /span> a, v, v0, x, x0; // Store data

= &nb= sp;

ifstream in(“physics.dat”); // Open the read file<= o:p>

= &nb= sp;

ofstream out(“newPhys.dat”); // Open the write file =

<= /o:p>

// Get the initial data=

in >= > v0 >> x0;

<= span style=3D'mso-spacerun:yes'> out << v0 << x0 <&l= t; endl;

<= /o:p>

// Calculates acceleration for eac= h line of data

while (in && !in.eof() = ) {

// Get the data

in >> v >> x;

<= /o:p>

// Calculate the accel= eration

a =3D v * v – v0 * v0 / ( 2 * (x * x – x0 * x0));

<= /o:p>

&n= bsp; // Output the result <= o:p>

<= /o:p>

out << v << x << a << endl;

<= /o:p>

}

<= /o:p>

// Close the files

<= /o:p>

in.clos= e();

out.clo= se();

<= o:p>

<= span style=3D'mso-spacerun:yes'> return(0);

<= /o:p>

Problem 3 (20 points): For this problem, you are given the files Card.h, Card.cpp, and main.cpp. Card.h is given below and declares the class Card. Add the following methods to the h= eader file:

isOneEyedJack, which takes no parameters and returns a Boolean value
containsSuit, which takes an array of Cards and the size of the array as parameters = and returns a Boolean value.

Modify the Card.cpp file = on the next page to define the method isOneEyedJack and the method c= ontainsSuit. Assume that the constructor, const= ants, and other methods are also defined in the .cpp file. You may use them when writing your code. In main you will add a test case to m= ore completely test the method containsSuit.

// file: Card.h

#include <string>

using std::string;

// Creates cards and provides a few method for the cards

#ifndef CARD_H

#define CARD_H

class Card {

private:

// Data Fields

int rank; // Numeric value of the card

int suit; // Suit of the card

public:

// Constants for names of ranks - example use Card::JACK

const static int JACK; = // value 11

const static int QUEEN; = // value 12

const static int KING; = // value 13

const static int ACE; = // value 14

// Constants for suits<= /span>

const static int CLUBS; = // value 0

const static int DIAMONDS; = // value 1

const static int HEARTS; = // value 2

const static int SPADES; = // value 3

// Creates a card

// @param r - rank of the card; @p= aram s - suit of the card

Card(int r, int s);

// Returns true exactly when the c= ard the method is invoked on is the

// the Jack of Spades or the Jack = of Hearts – a one-eyed Jack

bool isOneEyedJack();

// Returns true iff the hand (stor= ed in an array) contains a card of

// the same suit as the card the m= ethod is invoked on.

bool containsSuit(Char [], int);

// Returns the value of the rank

int getRank();

// Returns the value of the suit

int getSuit();

// Returns the name of the rank of= the card: 2, 3, …, King, Ace

string getRankName();

// Returns the name of the suit of= the card: Clubs, …, Spades

string getSuitName();

};

// file: Card.cpp

#include “Card.h” = &nb= sp; // Necessary file

// Returns true exactly when the card the method is invoked on is the

// the Jack of Spades or the Jack of Hearts – a one-eyed Jack=

bool Card::isOneEyedJack()

{

if (rank =3D=3D Card::JACK && (suit =3D=3D Card::SPADES

|| suit =3D=3D Card::HEARTS))

return true;

else<= o:p>

return false;

}

// Returns true iff the hand (stored in an array) contains a card of

// the same suit as the card the method is invoked on.

bool Card::containsSuit(Card [] hand, int size)

{

bool fo= und =3D false;

for (int i =3D 0= ; i < size; i++) {

if (hand[i].suit =3D=3D suit)

found =3D true;

}

return found;

}

// file: main.cpp

// Tests the methods in the card class to make sure the code works

// correctly. Add another test o= f the containsSuit method to more

// completely test it. For full credit, use constants.

#include <iostream>

using std::cout;

using std::endl;

#include = &nb= sp; // Necessary header file

int main()

{

Card jackS(Card::JACK, Card::SPADES);

Card queenS(Card::QUEEN, Car= d::SPADES);

Card hand[] =3D {Card(2, Card::HEARTS), Card(5, Card::HEARTS),

= Card(7, Card::CLUBS), Card(10, Card::DIAMONDS),

= queenS};

cout << jackS.isOneEyedJack() << endl;

cout << queenS.isOneEy= edJack() << endl;

cout << jackS.contains= Suit(hand, 5) << endl; =

// Add your test here

h= and[4] =3D Card(Card::QUEEN, Card::HEARTS);

cout <&= lt; jackS.containsSuit(hand, 5);

return 0;

}

Problem 4 (15 points): What is the output of the following program?

#include <iostream>

using std::cout; using std::endl;

#include “Card.h”

const int ASIZE =3D 5;

void mystery(Card[], int);

int main() {

Card hand[] =3D {Card(2, Card::HEA= RTS), Card(10, Card::HEARTS),

= Card(7, Card::CLUBS), Card(10, Card::SPADES),

= Card(6, Card::CLUBS)};

mystery(hand, ASIZE);

for (int i =3D 0; i < ASIZE; i+= +) {

cout << hand[i].getRankName() << " of "

= << hand[i].getSuitName() << ", ";

}

cout << endl;

}

void mystery(Card hand[], int size) {

if (size !=3D 0) {

Card c =3D hand[= 0];

for (int i =3D 1; i <= ; size; i++) {

if(c.getRank() < hand[i].getRank())

= c =3D hand[i];

else if (c.getRank() =3D=3D hand[i].getRank() &&

= c.getSuit() < hand[i].getSuit())

= c =3D hand[i];

}

int j =3D 0;

for (int k =3D 0= ; k < size-1; k++) {

if (c.getRank() =3D=3D hand[j].getRank() &&

= c.getSuit() =3D=3D hand[j].getSuit())

= j++;=

hand[k] =3D hand[j];

j++; =

}

cout << si= ze << ": ";

for (int i =3D 0= ; i < size-1; i++) {

cout << hand[i].getRankName() << " of "<= /p>

= << hand[i].getSuitName() << ", ";

}

cout << en= dl;

mystery(hand, si= ze-1);

hand[size-1] =3D= c;

}

5: 6 of Hearts, 10 of Hearts, 7 of Cl= ubs, 2 of Clubs,

4: 6 of Hearts, 7 of Clubs, 2 of Club= s,

3: 6 of Hearts, 2 of Clubs,

2: 2 of Clubs,

2 of Clubs, 6 of Hearts, 7 of Clubs, = 10 of Hearts, 10 of Spades,

Problem 5 (15= points): What is the output of this program? (Enter your answers at the end of this problem.)

#include <iostream>

using std::cout;

using std::endl;

class Circle

{

private:

int centerX;

int centerY;

int radius;

public:

// Constructors<= /p>

Circle();

Circle(int x, int y, int r);<= /o:p>

// Mutator

void setRadius(int r);<= /span>

// Accessors

int getRadius();=

int getDiameter();

};

// Constructor

Circle::Circle()

{

&n= bsp; centerX =3D 0;

centerY =3D 0;

radius =3D 1;

}

// Constructor

Circle::Circle(int x, int y, int r)

{

centerX =3D x;

centerY =3D y;

radius =3D r;

}

// Sets the radius

void Circle::setRadius(int r)

{

radius =3D r;

}

// Returns the radius

int Circle::getRadius()

{

return radius;

}

//Returns the diameters

int Circle::getDiameter()

{

return 2 * radius;

}

int main()

{

int x =3D 3;

int y =3D 6;

int r =3D 2;

Circle cir1;

Circle cir2(x, y, r);

cout << "a) "= ; << cir1.getDiameter() << endl;

cout << "b) "= ; << cir2.getDiameter() << endl;

y =3D 5;

Circle cir3(r, x, y);

cout << "c) "= ; << cir1.getRadius() << endl;

cout << "d) " << cir3.getDiameter() << endl;

cir2.setRadius(4);

cout << "e) "= ; << cir2.getDiameter() << endl;

cir1.setRadius(cir3.getDiameter());

cout << "f) " << cir1.getDiameter() << endl;

Circle *copy =3D &cir1;<= o:p>

&n= bsp; (*copy).setRadius(5);

cout << "g) "= ; << cir1.getRadius() << endl;

cout << "h) "= ; << cir3.getRadius() << endl;

cir3 =3D cir1;

cir3.setRadius(9);

cout << "i) " << cir1.getDiameter() << endl;

cout << "j) " << cir3.getDiameter() << endl;

copy =3D &cir2;

cout << "k) " << cir1.getDiameter() << endl;

cout << "l) " << cir2.getDiameter() << endl;

cout << "m) " << cir3.getDiameter() << endl;;

cout << "n) "= ; << (*copy).getDiameter() << endl;

}

Output: