Chapter 5

Iteration: Patterns for Classes and Programs

5.1

Here's one answer, there are many others.

    const int SENTINEL = 0;
    int count = 0;

    int num;
    int numNeg = 0;
    int numPos = 0;
    cin >> num;
    while(num != SENTINEL)
    {
	if (num > 0)
        {
	    numPos += 1;
	}
	else
        {
	    numNeg += 1;
	}
	count += 1;
	cin >> num;
    }
    cout << "number of positive #'s = " << numPos << endl;
    cout << "number of negative #'s = " << numNeg << endl;

5.2

Most systems will indicate 7 words have been read. The words are "This", "is", "the", "start," , "this", "is", "the".

5.3

const int FEET_PER_MILE = 5280;
const int OUNCES_PER_LB = 16;
const double E = 2.718;
const double GRAMS_PER_POUND = 453.59;
const double FT_LBS_PER_ERG =  1.356e7;

5.4

Giving word a value means the loop test will be true the first time it is checked, so the loop body will execute. This means an extraction cin >> ... statement is needed, then a check to see that the word extracted isn't the sentinel word.

    word = "dummy";
    while (word != LAST_WORD)
    {
        cin >> word;
        if (word != LAST_WORD)
        {
            numWords += 1;
        }
    }

5.5

Here's one solution:

    const String LAST_WORD = "end";
    const String LAST_WORD_II = "finish";

    int numWords = 0;
    String word = "dummy";
    while (word != LAST_WORD && word != LAST_WORD_II)
    {
        cin >> word;
        if (word != LAST_WORD && word != LAST_WORD_II)
        {
            numWords += 1;
        }
    }

5.6

Dos-machines and Unix machines support I/O redirection. Program 5.2 can be used on all machines that have C++ compilers. However, specifying a path to the directory containing data files may be more difficult on some machines than others. You may have a version of Program 5.4 (countw2.cc) that uses a "file-window" that allows you to navigate the file-system looking for files. There are 25,758 words in Romeo and Juliet

5.7

To modify Program 5.6 (numbers.cc) to read floating point numbers requires that the variables total and number be changed to type double. No other changes are necessary.

5.8

The statement

    cout << "average = " << (double) total/numNums << endl;

    cout << "average = " << double(total/numNums) << endl;

5.9

The cast

    int value = int(sqrt(25.3));

5.10

The following program will work:

#include 
#include 

main()
{
    cout << "largest int value = " << INT_MAX << endl;
    cout << "smallest int value = " << INT_MIN << endl;
}

In Program 5.7 (mindata.cc) changes needed to find the largest value read are:

• change variable name 'minimum' to 'maximum'
• initialize maximum to INT_MIN (accessible via )
• change message printed to reflect maximum value printed, no longer minimum

In Program 5.8 (mindata2.cc) the variable name must also change as must the message printed. In addition, the if statement inside the while loop should be

    while(cin >> number)
    {
        numNums += 1;
        if (number > maximum)
        {
            maximum = number;
        }
    }

Here's one program.

#include 

// compute sum of all odd and of all even numbers entered

main()
{
    const int SENTINEL = 0;    
    int num;
    int evenTotal = 0;
    int oddTotal = 0;
    int numOdd = 0;
    int numEven = 0;

    cin >> num;
    while (num != SENTINEL)
    {
        if (num % 2 == 0)           // even number, update stats
        {
            numEven += 1;
            evenTotal += num;
        }
        else
        {
            numOdd += 1;
            oddTotal += num;
        }
        cin >> num;
    }
    
    cout << "average of even numbers = " << double(evenTotal)/numEven << endl;
    cout << "average of odd numbers = " << double(oddTotal)/numOdd << endl;
}

5.13

To modify either Programs 5.2 or 5.3 to calculate the average word length in of the words processed requires adding a variable numLetters and a statement in the while loop to accumulate the total of all the letters.

    int numLetters = 0;
    while (infile >> word)
    {
        numWords += 1;
        numLetters += word.length();
    }
    cout << "average word length = " << double(numLetters)/numWords << endl;

    void ReadNums(int & numNums, int & total)
    // postcondition: returns number of numbers read (numNums)
    //                        and total of all numbers read (total)

The body of the function is the prompt and loop from Program 5.6. If only the average is to be calculated, the function header would be

    void ReadNums(double & average)

Alternatively, the value could be returned by the function, with no parameters passed into the function:

    double ReadNums()

    void GetName(String & first, String & last)
    // postcondition: return first/last names entered by user
    {
        cout << "enter first name: ";
        cin >> first;
        cout << "enter last name: ";
        cin >> last;
    }

It's possible to do all input on one line:

        cout << "enter first and last name (with space between): ";
        cin >> first >> last;

    int numLetters = 0;

    numLetters += word.Length();

can be used to update the number of letters read for each word. After the loop, the statement:

    average = double(numLetters)/(small + medium + large);

would be added. The header of the function requires that a double parameter 'average' be added:

void
FingerPrint(String filename, int & small, int & medium, int & large,
            double & average)

void
Roots(double a, double b, double c, double & root1, double & root2)
// precondition: a,b,c coefficients of ax^2 + bx + c     
// postcondition: sets root1 and root2 to roots of quadratic
// exception:     if roots are imaginary, set to infinity     
{
    double discriminant = b*b - 4*a*c;
    
    if (discriminant < 0)
    {
        root1 = root2 = infinity();         // from math.h
    }
    else                                    // get both roots
    {
        double base = sqrt(discriminant);
        root1 = (-b + base)/(2*a);
        root2 = (-b - base)/(2*a);
    }
}

     first += 2;

The statement

    second *= 2;

    void Change(int & first, int & second)
    {                   |      /
        first += 2;     |     /
        second *= 2;    |    /
    }                   |   /
                        |  /
    main()              | /
    {                +-----+
        int num = 8; |     |
                     +-----+
        Change(num,num);
        cout << num << endl;
    }

void ExtremeWords(String & firstAlph, String & lastAlph)
// postcondition: firstAlph = alphabetically first word read
//                lastAlph = alphabetically last word read     
{
	    // fence-post, first word read stored
    
    string word;
    if (cin >> word)
    {
        firstAlph = lastAlph = word;
    }

	 // read other words
    while (cin >> word)
    {
        if (word < firstAlph)
        {
	    firstAlph = word;
	}
	else if (lastAlph < word)
        {
	    lastAlph = word;
	}
    }
}

    void Swap(int & a, in t& b)
    // postcondition: inerchange values of a and b
    {
        int temp = a;      // need temporary storage
        a = b;
        b = temp;
    }    

    string response;
    do{
        cout << "enter response: ";
        cin >> response;
        if ("deposit" == response)
        {
            DoDeposit();
        }
        else if ("withdraw" == response)
        {
            DoWithdraw();
        }
        else if ("information" == response)
        {
            DoAccountInformation();
        }
    } while (response != "quit");

    int k = 0;
    while (k <= 100)
    {
	k += 2;
	cout << "even number = " << k << endl;
    }

is equivalent to:

    int k;
    for(k=0; k <= 100; k += 2)
    {
	cout << "even number = " << k << endl;
    }

The second loop can yield several different for loops, two are shown below.

    int total = 0;
    int num;
    while (cin >> num)
    {
	total += num;
    }

In this for loop, there are empty initialization and increment sections, but the loop is equivalent to the while loop above.

    int total = 0;
    int num;
    for(; cin >> num; )
    {
	total += num;
    }

In this for loop, the initialization and increment aren't really related to the guard at all, but the loop functions correctly. Try to avoid writing this kind of loop [although many programmers would like this style of programming.

    for(total = 0; cin >> num; total += num);    

Whenever it's possible for a loop to execute zero times, a while loop is preferable to a for loop. 5.25 A switch statement cannot be used in icecream.cc because in that program several if/else statements are used to distinguish among string values. A switch statement CANNOT have a string as its 'argument', only variables of type int (and char, and other integral types) can be used with switch statements. 5.26 See the primes program earlier in the book (primes.cc). 5.27

int numRows;
int j,k;

cout << "enter number of rows: ";
cin >> numRows;


// first pattern
// *
// * *
// * * *
// * * * *
// and so on

for(j=1; j <= numRows; j++)
{
    for(k=0; k < j; k++)      // iterate j times
    {
	cout << "* ";
    }
    cout << endl;
}


// for pattern below
//      *
//     * *
//    * * *
//   * * * *
//

const int MIDDLE = 40;        // middle of 80-column screen

for(j=1; j <= numRows; j++)
{
    // tab over
    for(k=0; k < MIDDLE - j + 1; k++)
    {
	cout << " ";
    }
    for(k=0; k < j; k++)
    {
	cout << "* ";
    }
    cout << endl;
}

   cout << "enter distance from start ";
   cin >> goal;
   do{
       jump = die.Roll();
       if (jump == 1)
       {
           position += 1;
       }
       else
       {
           position -= 1;
       }
   } while (abs(position) != goal);

   void DoSimulation(int numSteps, int & distFinal, int & distMax)
   // precondition: numSteps = duration of random walk simulation
   // postcondition: distFinal = final distance from start of simulated random walk
   //                        distMax = maximal distance from start

        position += 2*die.Roll() - 1;

since 2*die.Roll() is either 0 or 2, the right-hand-side of the expression evaluates to either -1 or +1. 5.31 To simulate a walk on a lattice (sometimes this kind of distance is called Manhattan-geometry after the gridwork-like layout of streets in New York City), a four sided die could be rolled with the outcome indicating whether a step is taken in the North/South/East/West direction. It is still necessary to keep both x/y coordinates.

Another approach might use a coin or two-sided die to determine whether the north/south OR the east/west direction would change. Another two sided die could determine up/down or left/right. 5.32 It is more likely that the one-dimensional random walk will visit the origin than that the two-dimensional walk will return to the origin. In both cases it is possible for this to happen. 5.33 Imagine a circle of n lilly-pads. If the pads are arranged in a circle, jumping from one to the other is like walking in a circle. This can be done by equating pads with numbers 0, 1, 2, ... (n-1). A "left" jump in the randwalk.cc corresponds to subtracting 1 from the current "pad". However, an if statement can be used to equate -1 with (n-1) so that such a jump continues around the circle.

   bool IsLeap(int year)
   // postcondition: returns true if year is a leap year, else false
   {
       Date d(2,1,year);         // February first of given year
       return d.LastDay == 29;
   }

    cout << d.DayOfWeek() << endl;

4 A field width of 4 will cause the days to be farther apart. Unfortunately, the header "S M T W Th F S" is NOT printed using a fieldwidth, so this will NOT be lined up properly.

Since dayOfWeek is incremented before the statement checking the remainder when divided by 7 ( dayOfWeek % 7 == 0 ) a new line is started after every Saturday (the comments in the code reflect this too). The if statement is needed after the loop in case the last week stopped in the middle of the week in which case the last line needs to be ended with a newline (endl).

5 Among other fencepost problems are

• printing a week: the first week is different than the other weeks because it has preceeding blank days.
• ending each week with a newline: all weeks except the last end with a Saturday.

6 To print a calendar for an entire year the function Calendar should be called 12 times, once for each month, i.e., as shown below (assuming that a function MonthName exists to print the name of a month given the corresponding number 1 == January, etc.).

    for(k=1; k <= 12; k+=1)
    {
        cout << MonthName(k) << endl;
        Calendar(k,year);
    }