Exercise 4-0

Compile, execute, and test the programs in this chapter

Solution

Chapter 4 (Organising programs and data) contains a mix of all learnings gained from previous chapters (0 to 3), with the newly added introduction to program partitioning – to break down a large program into multiple .cpp (source) files and .h (header) files to make it more manageable. In this post I will demonstrate my understanding of chapter 4 via presenting the one core project which encompasses the use of program partitioning.

My learning strategy:

1. Read through chapter 4 – try and understand as much as possible.
2. Write and execute the chapter 4 project in Code::Block – try and get the partitioned program work.
3. Read through chapter 4 again and experiment with the project – now that the project is working, I would like to understand why and how it works.
4. Document core learning outcome – this is what this post is for!

Now that I have spent 2 days completing (and re-iterating) step 1 to 3 above, I believe it is time to execute step 4 above – to document learning outcome via writing this post.

The Project

Purpose of the Chapter 4 project: to read in a flat-file format like input, and produce a summarised output (see diagram below).

Chapter 4 requires us to create a partitioned program (or so called project) that is formed of multiple .cpp source files and .h header files. These files somewhat “know” about each other and can work together in the solving of big problem.

In Code::Block (or probably most of the mainstream IDE), we can create such a project fairly easily to keep these files tidy / organised.

I now document the C++ Source Files and Header Files in the following sections.

C++ Source Files

• main.cpp – this is the first program that is run during the implementation phase.
• grade.cpp – contains all functions relating to computing grades.
• median.cpp – contains all functions relating to computing median.
• Student_info.cpp – contains all functions relating to handling a Student_info object.

C++ Header Files

• grade.h – declare the functions as defined in grade.cpp
• median.h – declare the functions as defined in median.cpp
• Student_info.h – declare the functions as defined in Student_info.cpp, plus defining the data structure of the Student_info (object) type.

See my post How to add header and source files in Code::Block for additional information.

This diagram below shows what the Code::Block Management Tree look like after successful creation of these files.

The actual content of the source and header files are documented in the following sections.

Source Files

main.cpp

#include <algorithm>
#include <iomanip>
#include <ios>
#include <iostream>
#include <stdexcept>
#include <string>
#include <vector>
#include "grade.h"
#include "Student_info.h"

using std::cin;
using std::cout;
using std::endl;
using std::domain_error;
using std::max;
using std::setprecision;
using std::sort;
using std::streamsize;
using std::string;
using std::vector;

int main()
{
    vector<Student_info> students;
    Student_info record;
    string::size_type maxlen = 0;   // the length of the longest name

    // read and store all the student's data.
    // Invariant:   students contain all the student records read so far
    //              maxlen contains the length of the longest name in students
    while (read(cin, record))
    {
        // find the length of longest name
        maxlen = max(maxlen, record.name.size());
        students.push_back(record);
    }

    // alphabetize the student records
    sort(students.begin(), students.end(), compare);

    // write the names and grades
    for (vector<Student_info>::size_type i = 0;
         i != students.size(); ++i)
    {
        //write the name, padded on teh right to maxlen + 1 characters
        cout << students[i].name
             << string(maxlen + 1 - students[i].name.size(), ' ');

         //compute and write the grade
        try
        {
            double final_grade = grade(students[i]);
            streamsize prec = cout.precision();
            cout << setprecision(3) << final_grade
                 << setprecision(prec);
        }
        catch (domain_error e)
        {
            cout << e.what();
        }
        cout << endl;
    }
    return 0;
}

grade.cpp

#include <stdexcept>
#include <vector>
#include "grade.h"
#include "median.h"
#include "Student_info.h"

using std::domain_error;
using std::vector;

// definitions for the grade functions from S4.1/52, S4.1.2/54, S4.2.2/63

// compute a student's overall grade from midterm and final exam
// grades and homework grade (S4.1/52)
double grade(double midterm, double final, double homework)
{
    return 0.2 * midterm + 0.4 * final + 0.4 * homework;
}

// compute a student's overall grade from midterm and final exam grades
// and vector of homework grades.
// this function does not copy its argument, because median (function) does it for us.
// (S4.1.2/54)
double grade(double midterm, double final, const vector<double>& hw)
{
    if (hw.size() == 0)
        throw domain_error("student has done no homework");
    return grade(midterm, final, median(hw));
}

// this function computes the final grade for a Student_info object
// (S4.2.2/63)
double grade(const Student_info& s)
{
    return grade(s.midterm, s.final, s.homework);
}

median.cpp

// source file for the median function
#include <algorithm>
#include <stdexcept>
#include <vector>

using std::domain_error;
using std::sort;
using std::vector;

// compute the median of a vector<double>
double median(vector<double> vec)
{
    typedef vector<double>::size_type vec_sz;

    vec_sz size = vec.size();
    if (size == 0)
        throw domain_error("median of an empty vector");

    sort(vec.begin(),vec.end());

    vec_sz mid = size/2;

    return size % 2 == 0 ? (vec[mid] + vec[mid-1]) / 2 : vec[mid];
}

Student_info.cpp

#include "Student_info.h"

using std::istream;
using std::vector;

// we are interested in sorting the Student_info object by the student's name
bool compare(const Student_info& x, const Student_info& y)
{
    return x.name < y.name;
}

// read student's name, midterm exam grade, final exam grade, and homework grades
// and store into the Student_info object
// (as defined in S4.2.2/62)
istream& read(istream& is, Student_info& s)
{
    // read and store the student's name and midterm and final exam grades
    is >> s.name >> s.midterm >> s.final;

    // read and store all the student's homework grades
    read_hw(is, s.homework);
    return is;
}

// read homework grades from an input stream into a vector<double>
// (as defined in S4.1.3/57)
istream& read_hw(istream& in, vector<double>& hw)
{
    if (in)
    {
        // get rid of previous contents
        hw.clear();

        // read homework grades
        double x;
        while (in >> x)
            hw.push_back(x);

        // clear the stream so that input will work for the next student
        in.clear();
    }
    return in;
}

Header Files

grade.h

#ifndef GUARD_GRADE_H
#define GUARD_GRADE_H

//grade.h
#include <vector>
#include "Student_info.h"

double grade(double, double, double);
double grade(double, double, const std::vector<double>&);
double grade(const Student_info&);

#endif // GUARD_GRADE_H

median.h

#ifndef GUARD_MEDIAN_H
#define GUARD_MEDIAN_H

// median.h - final version
#include <vector>
double median(std::vector<double>);

#endif // GUARD_MEDIAN_H

Student_info.h

#ifndef GUARD_STUDENT_INFO_H
#define GUARD_STUDENT_INFO_H

// Student_info.h
#include <iostream>
#include <string>
#include <vector>

struct Student_info
{
    std::string name;
    double midterm, final;
    std::vector<double> homework;
};

bool compare(const Student_info&, const Student_info&);
std::istream& read(std::istream&, Student_info&);
std::istream& read_hw(std::istream&, std::vector<double>&);

#endif // GUARD_STUDENT_INFO_H

Test Program

I believe that by test running the program multiple times (and differently each time) it will enable me to understand a bit more about why and how the program works as a whole. Experiment, experiment, and experiment…

After compiling all the files followed by hitting the run program button, a blank command window fires up awaits me to provides input. I performed the various tests using different input values (or format). The results will hopefully enable me to visualise patterns and understand the program a bit more.

Test 1

I will now input all values in 1 line, hit enter, then hit end-of-file (F6), then hit end-of-file (F6) again. See what the output looks like and why it appears that way.

Test 1 – Input and Result

Johnny 70 80 50 60 30 Fred 95 90 100 100 100 Joe 40 40 50 60 50 70 70 50
^Z
^Z
Fred   95
Joe    46
Johnny 66

Process returned 0 (0x0)   execution time : 121.246 s
Press any key to continue.

Test 1 – Observation and Explanation

1. The first while (read(cin, record)) { } (within the main program) activates the std::cin which enables user the type-in input values via the console window.
2. I type all values in one line (separated by a space character), like this: name, midterm score, final score, homework scores.
3. I then hit the enter button to open up a new line. This “hitting the enter button” action parses the values that I typed, into a buffer.
4. The istream& read(istream&, Student_info&) function (as defined in Student_info.cpp) parse the first buffer value “Johnny” to s.name, and clear that value from the buffer.
5. It then parse the (now first) buffer value 70 to s.midterm, and clear that value from the buffer.
6. It then parse the (now first) buffer value 80 to s.final, and clear that value from the buffer.
7. The istream& read_hw(istream&, vector<double>&) function is then invoked (as defined in Student_info.cpp). It prepares an empty vector<double>& hw. The while (in >> x) parses all the valid values from the buffer to the (vector) hw, until the value become invalid (e.g. a string rather than a number). In this case, this procedure parses the 50, 60, and 30 to hw[0], hw[1], hw[2] respectively. When the procedure encounters the (non-numeric) value “Fred”, it exits the while automatically and change the status of the istream& in to an error status. The in.clear() reset the error status to enable smooth data parse for the next student. Because the “Fred” was not parsed during this while loop (as the while loop got exited due to non-numeric value), and therefore not cleared from the buffer, it now becomes the first value of the buffer (this is an important note to make – because in the 2nd loop, the program now able to parse “Fred” as a name of the 2nd Student_info object!).
8. The while (read(cin, record)) { } then enters the 2nd loop (to process Fred’s scores). It then repeats in the 3rd loop to process Joe’s scores. In the end, the vector<Student_info> students contains the 3 Student_info objects (i.e. “Johnny”, “Fred”, and “Joe”)
9. After processing the entirety of the one-liner input, I enter end-of-file button. This has the effect of exiting the while loop of read_hw function.
10. I enter the enter end-of-file button once more time. This has the effect of exiting the while loop (of the main program).
11. Now that both loops are exited, the main program then proceeds to the sort(students.begin(), students.end(), compare) phase. The downstream block of code output the result in a nicely formatted summary showing the overall score for each student, sorted by the student’s name.

The above explanation is not comprehensive, as to explain the whole program, it would take multiple pages! The main reason that I decided to document the above is to highlight these core observations / concepts:

• The behaviour of std::cin and buffer – my previous post Solution to Exercise 1-6 has enabled me to make sense of why and how this chapter 4 program works. e.g. the effect of hitting that enter button first time round!
• The first end-of-file exits the inner-most while loop (the one within the read_hw function).
• The second end-of-file exits the outer-most while loop (the one within the main program) – which enable the implementation to continue to the sort step (within the main program).
• Chapter 4 of the book has explained most of the details in depth – so I am not going to repeat here.

Test 2

I will now input in the most consistent and most understandable format. i.e. input the values 1 line per student (Name, mid-term score, final score, and homework scores). When I am done I will hit enter, then hit end-of-file (F6), then hit end-of-file (F6) again. See what the output looks like and why it appears that way. (I will use the same values as of test 1 – to hopefully prove that the output result would be the same as test 1.)

Test 2 – Input and Result

Johnny 70 80 50 60 30
Fred 95 90 100 100 100
Joe 40 40 50 60 50 70 70 50
^Z
^Z
Fred   95
Joe    46
Johnny 66

Process returned 0 (0x0)   execution time : 33.322 s
Press any key to continue.

Test 2 – Observation and Explanation

The output of this test is exactly the same as test 1. This is not surprising. The overall process of test 2 is mostly similar to test 1, with one very minor difference: in test 1 we input all the values in 1 line and hit enter – this parses all values (for all 3 students) in the buffer. The downstream process then read from the buffer and proceed accordingly, and eventually created the 3 Student_info type objects.

In this test 2, we input the values for student 1 in 1 line. Hitting enter parse the values of this 1 student into the buffer. The read() function reads the name, then mid-term score, then the final score, then the read_hw function (within the read function) reads the vector elements homework 0 to home work 4. The implementation then awaits for our next homework score.

Then we type the values for the 2nd student (Fred) in a similar fashion. This time, after we hit the enter button to open up the 3rd line, the read_hw function (that we talked about just now) that is expecting a numeric homework 5, suddenly “sees” this non-numeric (string) value “Fred”. It exit the while loop (of the read_hw function), create an error status, then clear that error status as per the in.clear() (to enable smooth read of the next student). Going back to the main program, the 2nd while loop while (read(cin, record)) { } start reading that “Fred” (first value of the buffer) as the student’s name, followed by reading the renaming numeric values in the buffer (as mid-term score, final-score, homework scores). This cycle repeats for the 3rd student “Joe”.

Like test 1, the first end-of-file (F6) button exit the inner while loop (of the read_hw function). The second end-of-file (F6) button exit the outer while loop (of the main program).

The main program then proceeds with the downstream block of code, and output the results accordingly.

Test 3

This time in test 3, I combine a bit of test 1 and test 2 together. i.e. I will use the same set of values, but this time, some of these values shall spread over multiple lines, and some on the same line. I would like to prove that the result should be exactly the same as test 1 and 2, using the hybrid explanations as per test 1 and test 2.

Test 3 – Input and Result

Johnny 70 80 50
60 30
Fred 95 90 100 100 100
Joe
40
40
50
60 50 70 70 50
^Z
^Z
Fred   95
Joe    46
Johnny 66

Process returned 0 (0x0)   execution time : 47.050 s
Press any key to continue.

Test 3 – Observation and Explanation

As expect, the result is exactly the same as test 1 and test 2. This has proved that, using the explanation as per test 1 and test 2, as long as the input values are the same, it doesn’t matter whether we spread our data over multiple lines or on the same line. However, I do find the input format of test 2 (i.e. one line per student) is the most tidy and easy-to-understand flat-file format. In fact, most of the flat-files that I deal with at work (such as reading CSV files using SAS) likes this type of format – 1 line per observation (or record). So my recommendation is to stick with the (CSV like) flat-file format used in test 2.

Test 4

In this test, I would like to demonstrate what the result looks like, if I enter no homework for some students.

Test 4 – Input and Result

Johnny 70 80 50 60 30
Leon 100 100
Fred 95 90 100 100 100
Simon 90 90
Joe 40 40 50 60 50 70 70 50
^Z
^Z
Fred   95
Joe    46
Johnny 66
Leon   student has done no homework
Simon  student has done no homework

Process returned 0 (0x0)   execution time : 48.859 s
Press any key to continue.

Test 4 – Observation and Explanation

Note that Leon and Simon have done no homework! And as expected, the program is clever enough to pick this up and store this status for the corresponding Student_info type objects, instead of exiting the program entirely.

This “exception handling” step is carried out during the main.cpp program, between the try and catch exception handling step.

Test 5

This time, I enter 2 lines of input correctly. Then for the 3rd line, I only enter a name (and then hit enter). Then on the 4th line, I enter another name and hit enter. This time the program only processes the first two lines of input and output the results for these two lines. The program ignore the 3rd and 4th invalid lines entirely. This is as expected.

Test 5 – Input and Result

Johnny 70 80 50 60 30
Fred 95 90 100 100 100
Joe
Simon
Fred   95
Johnny 66

Process returned 0 (0x0)   execution time : 29.578 s
Press any key to continue.

Test 5 – Observation and Explanation

On the 3rd line, after entering “Joe, the read function (within the Student_info.cpp file) expects a numeric value midterm score. Because the next value “Simon” is not a numeric value, it exit that loop and return an error status (throug the lvalue istream& is). Because of this, the while loop within the main program exits, and proceeds with the downstream process. Also, because the 3rd line never make it to the read_hw function phase, the 3rd Student_info type object was never created. i.e. only the object for student Johnny and Fred were created. Hence the output result only contains these two students.

Conclusion

These tests conclude that the program works as long as the input data is in a consistent and expected format. The flat-file format as per test 2 is probably the best one to use (i.e. one line per record) – as it is easy to understand and consistent. Chapter 4 has really taught me a great deal on partitioning a program, and refreshing me the way std::cin and buffer function. Data extract-transform-load (ETL) is a core process used in industry reading flat-files. This chapter 4 has helped me understanding how C++ handle ETL.

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

Exercise 3-6

The average-grade computation in §3.1/36 might divide by zero if the student didn’t enter any grades. Division by zero is undefined in C++, which means that the implementation is permitted to do anything it likes. What does your C++ implementation do in this case? Rewrite the program so that its behavior does not depend on how the implementation treats division by zero.

Solution

Before changing any codes, let me try running the program like this:

• Test 1: do not supply any grades at all.
• Test 2: supply mid-term and final-exam grades, but not supply the homework grades.

Test 1

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: ^Z
Enter all your homework grades, followed by end-of-file: Your final grade is nan

Test 2

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: 80
90
Enter all your homework grades, followed by end-of-file: ^Z
Your final grade is nan

In both tests, because of the fact that I did not specify any homework grades, when the program tries to compute the average homework grade, it bump into division by zero. In my case, I ran the test via the Code::Block IDE on a Windows Vista machine, the undefined output is nan. (i.e. Not A Number).

To avoid this add an if statement to condition-check the value of count. If count is zero, exit the code peacefully with a message saying “Cannot compute final grade due to missing grades supplied – ensure you supply all grades as required.” Something like that.

 if (count == 0)
 {
 cout << "Cannot compute final grade due to missing grades supplied - ensure you supply all grades as required." << endl;
 return 1;
 }

The full code looks like this:

#include <iomanip>
#include <ios>
#include <iostream>
#include <string>

using std::cin;
using std::cout;
using std::endl;
using std::setprecision;
using std::string;
using std::streamsize;

int main()
{
    // ask for and read the student's name
    cout << "Please enter your first name: ";
    string name;
    cin >> name;
    cout << "Hello, " << name << "!" << endl;

    // ask for and read the midterm and final grades
    cout << "Please enter your midterm and final exam grades: ";
    double midterm, final;
    cin >> midterm >> final;

    // ask for the homework grades
    cout << "Enter all your homework grades, "
            "followed by end-of-file: ";

    // the number and sum of grades read so far
    int count = 0 ;
    double sum = 0.0;

    // a variable into which to read
    double x;

    // invariant:
    //    we have read count grades so far, and
    //    sum is the sum of the first count grades
    //    after entering the last value, hit the F6 button, then enter (to indicate end of file)
    //    or hit Ctrl+z, then enter.
    while (cin >> x)
    {
        ++count;
        sum += x;
    }

    double dummy = count; // for some reason the code fails unless I add this line.

    if (count == 0)
    {
        cout << "Cannot compute final grade due to missing grades supplied - ensure you supply all grades as required." << endl;
        return 1;
    }


    // write the result
    streamsize prec = cout.precision();

     cout << "Your final grade is " << setprecision(3)
         << 0.2 * midterm + 0.4 * final + 0.4 * sum / count
         << setprecision(prec) << endl;

    return 0;

}

 

Result

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: ^Z
Enter all your homework grades, followed by end-of-file: Cannot compute final gr
ade due to missing grades supplied - ensure you supply all grades as required.

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

Exercise 3-5

Write a program that will keep track of grades for several students at once. The program could keep two vectors in sync. The first should hold the student’s names, and the second the final grades that can be computed as input is read. For now, you should assume a fixed number of homework grades. We’ll see in §4.1.3/56 how to handle a variable number of grades intermixed with student names.

Solution

This exercise is a good one. The nature of the problem and the eventual solution builds on all the fundamentals that we have learnt from this chapter. There are many ways to solve this. Here is one of these many ways.

The solution strategy:

• Create a constant numHomework that defines the fix number of homework scores required for each student.
• Have an infinite while loop to enable user to enter the studentName – there will be points where the user can exit the program peacefully by entering the end-of-file key (Ctrl-Z / F6 for windows).
• As soon as a studentName is read, we append it to the studentNames vector – this vector store the studentName elements.
• Within the while loop we have a for loop to enable user to enter the homeworkScore one by one- until the predefined constant numHomework is reached.
• During the process we compute the totalScore and subsequently the meanScore for the corresponding student.
• As soon as the meanScore is computed, we append it to the meanScores vector – this vector store the list of meanScore elements.
• The nature of the looping systems ensure the two vectors studentNames and meanScores are always in sync.
• Allow user to either continue (by entering another studentName), or exit (by entering the end-of-file key. i.e. Ctrl-Z or F6 for windows).
• Output the pairs of studentName and meanScore elements of the vectors studentNames and meanScores using a for loop.

Putting this all together, we have our full program.

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <ios>
#include <string>
#include <vector>

using std::cin;             // <iostream>
using std::cout;            // <iostream>
using std::endl;            // <iostream>
using std::setprecision;    // <iomanip>
using std::sort;            // <algorithm>
using std::streamsize;      // <ios>
using std::string;          // <string>
using std::vector;          // <string>


int main()
{
    typedef vector<double>::size_type vecSize;

    const vecSize numHomework = 5;     // max number of homework per student

    string studentName;
    vector<string> studentNames;

    double homeworkScore;
    double totalScore;
    double meanScore;
    vector<double> meanScores;

    cout << "Enter student name: ";
    while (cin >> studentName)
    {
        studentNames.push_back(studentName);
        cout << "Enter " << numHomework << " homework scores below..." << endl;

        totalScore = 0; // Initialise
        meanScore = 0;  // Initialise

        for (vecSize i = 0; i != numHomework ; ++i)
        {
            cin >> homeworkScore;
            totalScore += homeworkScore;
        }

        meanScore = totalScore / numHomework;
        meanScores.push_back(meanScore);

        cout << "Enter another student name "
                "(or enter F6 key to exit): ";
    }

    vecSize numStudents = studentNames.size();
    cout << endl;
    cout << "Number of students entered: " << numStudents << endl;

    streamsize prec = cout.precision();
    for (vecSize i = 0; i != numStudents ; ++i)
    {
        cout << endl;
        cout << "Student: " << studentNames[i] << endl;
        cout << "Mean Score: " << setprecision(5)
            << meanScores[i] << setprecision(prec) << endl;
    }

    return 0;

}

Result

I now run a test to demonstrate the program output.

Enter student name: Johnny
Enter 5 homework scores below...
40
50
60
70
80
Enter another student name (or enter F6 key to exit): Fred
Enter 5 homework scores below...
99
78
67.5
66.8
100
Enter another student name (or enter F6 key to exit): Joe
Enter 5 homework scores below...
0
0
10
2
0
Enter another student name (or enter F6 key to exit): ^Z

Number of students entered: 3

Student: Johnny
Mean Score: 60

Student: Fred
Mean Score: 82.26

Student: Joe
Mean Score: 2.4

Process returned 0 (0x0)   execution time : 55.152 s

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

Exercise 3-4

Write a program to report the length of the longest and shortest string in its input.

Solution

I notice that the solution to this problem will be somewhat similar to the Solution to Exercise 3-3 (identify number of distinct elements in a vector). Except this time we are determining the length of the longest and shortest (string) elements. I therefore expect the eventual program to be similar.

Strategy

• Ask user to provide the list of words so we can append the string elements to a vector, v.
• Compute the size of vector, N.
• If vector size is 0, output an error message (as we need at least 1 string). Exit program peacefully with return code 1.
• If vector size is 1, we compute the size of the string element v[0]. The length of the longest string, SL, will be the same as the length of the shortest string, SS.
• If vector size is 2 or above, we apply an algorithm to compute SL and SS (to be explained further below).

Algorithm

The algorithm (for the case of vector size of 2 or above) is described as followings:

• Notice that this time we do not need to sort the vector – the condition checks downstream will be sufficient to update SS and SL.
• We initialise SL and SS to the size of the first string element v[0]. SL and SS will be updated accordingly during the comparison step.
• We initialise the index B to 1. This will be used for comparing the length of the v[B] with respect to SS and SL.
• We perform N-1 number of comparisons between v[B] and the current SL and SS. (e.g. if there are 10 elements, we can only compare 9 sets of adjacent elements).
• During each comparison step, if v[B].size() is larger than SL, we re-assign SL to v[B].size().
• If v[B].size() is smaller than SS, we re-assign SS to v[B].size().
• (If the two element lengths are equal, we do nothing.)
• We then increment the indices B by 1 for the next comparisons.
• We display the final value of SS and SL – these are the lengths of the shortest and longest strings elements within the vector v.

The Program

Putting this all together, we have our full program.

#include <iostream>
#include <string>
#include <vector>

using std::cin;             // <iostream>
using std::cout;            // <iostream>
using std::endl;            // <iostream>
using std::string;          // <string>
using std::vector;          // <string>

int main()
{
    // display header message
    cout << "***************************************************************\n"
            "*** This program reports the longest and shortest strings   ***\n"
            "***************************************************************\n";
    cout << endl;

    // ask for a list of numbers and store the list as a vector
    cout << "Enter a list of words one by one: ";
    vector<string> v;
    string x;
    while (cin >> x)
        v.push_back(x);    // append new input to the vector

    cout << endl;

    // define and compute core vector variables
    typedef vector<string>::size_type vecSize;   // define a type for vector size related variables
    vecSize N = v.size();            // number of elements in the vector
    vecSize numLoops = N - 1;        // number of (comparison) operators required

    typedef string::size_type strSize;   // define a type for string size related variables
    strSize SL;              // the length of the longest word
    strSize SS;              // the length of the shortest word

    // Check vector size, action accordingly
    if (N ==0 )
    {
        cout << "You need to enter at least 1 word! " << endl;
        return 1;
    }

    else if (N ==1 )
    {
        SL = v[0].size();
        cout << "Only 1 string supplied. The length of string = " << SL << endl;
        return 0;
    }

    else
    {
        // display some results to console window
        cout << "Vector size (number of words entered): " << N << endl;
        cout << endl;

        // declare new variables
        vecSize A = 0;     // vector index
        vecSize B = 1;     // vector index
        SS = v[0].size();            // the length of the shortest word
        SL = v[0].size();            // the length of the longest word

        // Loop through the vector, compute ND, and compute SS and SL
        for (vecSize i = 0; i != numLoops; ++i)
        {
            if (v[B].size() > SL)
            {
                SL = v[B].size();
            }
            if (v[B].size() < SS)
            {
                SS = v[B].size();
            }
            ++B;
        }
        // Display final results
        cout << endl;
        cout << "Length of shortest word: " << SS << endl;
        cout << "Length of longest word: " << SL << endl;
    }
    return 0;
}

Result

Below shows the results of the 3 sets of test:

• N = 0
• N=1
• N>=2

The results appear to agree well with the algorithm used.

N = 0

***************************************************************
*** This program reports the longest and shortest strings   ***
***************************************************************

Enter a list of words one by one: ^Z

You need to enter at least 1 word!

Process returned 1 (0x1)   execution time : 3.156 s

N = 1

***************************************************************
*** This program reports the longest and shortest strings   ***
***************************************************************

Enter a list of words one by one: a23456789
^Z

Only 1 string supplied. The length of string = 9

N >= 2

***************************************************************
*** This program reports the longest and shortest strings   ***
***************************************************************

Enter a list of words one by one: a23456789
a23456
a2345b
a234
a234567890
^Z

Vector size (number of words entered): 5


Length of shortest word: 4
Length of longest word: 10

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

Exercise 3-0

Compile, execute, and test the programs in this chapter.

Solution

There are two programs in this chapter. Both computes the overall student’s grade. The only difference resides on the algorithm used – one uses average homework grades, and the other one uses median homework grades.

Program 1 algorithm uses average homework grades

OverallGrade = (0.2 * midterm) + (0.4 * final) + (0.4 * sum / count)

Program 2 algorithm uses median homework grades.

OverallGrade = (0.2 * midterm) + (0.4 * final) + (0.4 * median)

The two programs are fully documented in chapter 3 of the book. The aim of this post is purely to test run these programs to ensure that they run as expected.

These are the two programs:

Program 1 (uses average homework grades)

#include <iomanip>
#include <ios>
#include <iostream>
#include <string>

using std::cin;
using std::cout;
using std::endl;
using std::setprecision;
using std::string;
using std::streamsize;

int main()
{
    // ask for and read the student's name
    cout << "Please enter your first name: ";
    string name;
    cin >> name;
    cout << "Hello, " << name << "!" << endl;

    // ask for and read the midterm and final grades
    cout << "Please enter your midterm and final exam grades: ";
    double midterm, final;
    cin >> midterm >> final;

    // ask for the homework grades
    cout << "Enter all your homework grades, "
            "followed by end-of-file: ";

    // the number and sum of grades read so far
    int count = 0 ;
    double sum = 0.0;

    // a variable into which to read
    double x;

    // invariant:
    //    we have read count grades so far, and
    //    sum is the sum of the first count grades
    //    after entering the last value, hit the F6 button, then enter (to indicate end of file)
    //    or hit Ctrl+z, then enter.
    while (cin >> x)
    {
        ++count;
        sum += x;
    }

    double dummy = count; // for some reason the code fails unless I add this line.

    // write the result
    streamsize prec = cout.precision();

     cout << "Your final grade is " << setprecision(3)
         << 0.2 * midterm + 0.4 * final + 0.4 * sum / count
         << setprecision(prec) << endl;

    return 0;
}

Note that I added this line after the while loop:

    double dummy = count; // for some reason the code fails unless I add this line.

For some unknown reason, by not including this line here, the code generate this toxic error following compilation:

Process returned 1968961623 (0x755BF857)

After I have added that “dummy” line to the code it then runs fine. I have no idea why though! (I run this via the Code::Block IDE on a Window Vista machine)

I will publish the test result at the Result section at the bottom of the post.

Program 2 (uses median homework grades)

#include <algorithm>
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <vector>

using std::cin;
using std::cout;
using std::endl;
using std::setprecision;
using std::sort;
using std::streamsize;
using std::string;
using std::vector;

int main()
{
    // ask for and read the student's name
    cout << "Please enter your first name: ";
    string name;
    cin >> name;
    cout << "Hello, " << name << "!" << endl;

    // ask for and read the midterm and final grades
    cout << "Please enter your midterm and final exam grades: ";
    double midterm, final;
    cin >> midterm >> final;

    // ask for the homework grades
    cout << "Enter all your homework grades, "
            "followed by end-of-file: ";

    vector<double> homework;
    double x;

    // invariant: homework contains all the homework grades read so far
    while (cin >> x)
        homework.push_back(x);

    // check that the student entered some homework grades
    typedef vector<double>::size_type vec_sz;
    vec_sz size = homework.size();
    if (size == 0)
    {
        cout << endl << "You must enter your grades. "
                        "Please try again." << endl;
        return 1;
    }

    // sort the grades
    sort(homework.begin(), homework.end());

    // compute the median homework grade
    vec_sz mid = size/2;
    double median;
    median = size % 2 == 0 ? (homework[mid] + homework[mid-1]) / 2
                           : homework[mid];

    // compute write the final grade
    streamsize prec = cout.precision();
    cout << "Your final grade is " << setprecision(3)
         << 0.2 * midterm + 0.4 * final + 0.4 * median
         << setprecision(prec) << endl;

    return 0;
}

This program compile and execute ok.

Note that the end-of-file key is Ctrl-Z (or F6) for windows, and Ctrl-D for Linux / Unix. (or google for the answer!).

i.e. after typing the last number and hit enter, hit the end-of-file key followed by another enter – that will indicate end-of-file.

Result

I will now run 4 tests and compare results with Excel.

	Even number of homework grades	Odd number of homework grades
Program 1 (uses average homework grades)	Test 1-even	Test 1-odd
Program 2 (uses median homework grades)	Test 2-even	Test 2-odd

These are the test values I will be using, along with the average and median figures computed in Excel.

	Even number of homework grades	Odd number of homework grades
Midterm Grade	70	70
Final Grade	80	80
Homework Grades	2.00	2.00
	5.00	5.00
	30.00	30.20
	40.66	45.50
	50.00	66.66
	60.00
Average Homework Grade	58.51 (Compare with Test 1-even)	57.97 (Compare with Test 1-odd)
Median Homework Grade	60.13 (Compare with Test 2-even	58.20 (Compare with Test 2-odd)

Test 1-even Result

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: 70.00
80.00
Enter all your homework grades, followed by end-of-file: 2.00
5.00
30.00
40.66
50.00
60.00
^Z
Your final grade is 58.5

The result agrees with Excel (58.51) up to 3 significant figure.

Test 1-odd Result

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: 70.00
80.00
Enter all your homework grades, followed by end-of-file: 2.00
5.00
30.50
45.50
66.66
^Z
Your final grade is 58

The result agrees with Excel (57.97) up to 3 significant figure.

Test 2-even Result

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: 70.00
80.00
Enter all your homework grades, followed by end-of-file: 2.00
5.00
30.00
40.66
50.00
60.00
^Z
Your final grade is 60.1

The result agrees with Excel (60.13) up to 3 significant figure.

Test 2-odd Result

Please enter your first name: Johnny
Hello, Johnny!
Please enter your midterm and final exam grades: 70.00
80.00
Enter all your homework grades, followed by end-of-file: 2.00
5.00
30.50
45.50
66.66
^Z
Your final grade is 58.2

The result agrees with Excel (58.20) up to 3 significant figure.

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

Exercise 2-10

Explain each of the uses of std:: in the following program:

int main ()
{
    int k = 0;
    while (k != 10)   // invariant: we have written k asterisks so far
    {
        using std::cout;
        cout << "*";
        ++k;
    }
    std::cout << std::endl;   // std:: is required here
    return 0;
}

Solution

Back to basics – what are namespace, names within a namespace, standard library, and iostream header? (refer to page 3 / chapter 0.5 of the book).

• std:: corresponds to a namespace called std.
• A namespace (e.g. std) is a collection of related names (e.g. cout, cin, endl). The standard library uses std to contain all the names that it defines.
• The iostream standard header defines the names (e.g. cout, cin, endl)

i.e. the iostream standard header defines the names cout, cin, endl, and we refer to these names as std::cout, std::cin, std::endl.

Now, let’s explain the uses of std:: in the program.

Within the while loop, we have using std::cout statement to declare that the name cout (residing in namespace std) is made available to the program. i.e. within the while loop scope whenever we have cout, the program “knows” that we are referring to std::cout – the cout of namespace std. The usage of std::cout statement has the life span of only within the while loop scope, as defined by the two curly braces { }. The while loop uses the std::cout to write asterisks on the console window.

Immediately after the while loop scope, we have std::cout – i.e. we need to explicitly specify that we are referring to the cout of the namespace std (and not other namespaces). std::cout is used here to write to the console window.

Lastly, we have the std::endl. Again, we need to explicitly specify that we are referring to the endl of the namespace std (and not other namespaces). std::endl is used for ending the currently line (and start a new line).

Putting this all together, the program does this:

• Write 10 asterisks to the console output window.
• End the current line and start a new line.

To run the program however, we must add the iostream header at the top which defines what the like of std::cout, std::endl, etc.

#include <iostream>

Result

**********

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

Exercise 2-9

Write a program that asks the user to enter two numbers and tells the user which number is larger than the other.

Solution

Here is my solution strategy:

• Obtain the two numbers from user. i.e. a and b.
• Perform a condition check. There are 3 possible cases.
• Case 1: a == b
• Case 2: a > b
• Case 3: a < b
• Prepare the outcome of each case. e.g. each case can have its own output message.

Putting this together, we have our full program.

#include <iostream>
using std::cout;
using std::cin;
using std::endl;

int main()
{
    cout << "***********************************************************************\n"
         << "*** This program compare 2 numbers and tell you which one is larger ***\n"
         << "***********************************************************************\n";
    cout << endl;
    cout << "Enter the first number (a): ";
    int a;
    cin >> a;
    cout << "Enter the second number (b): ";
    int b;
    cin >> b;
    cout << "a = " << a << ", b = " << b << endl;
    if (a == b)
        {cout << "a == b";}
    else if (a > b)
        {cout << "a > b";}
    else
        {cout << "a < b";}
    return 0;
}

Result

Let’s supply 3 sets of input (representing the 3 cases) and confirm the condition check works.

Case 1: a == b

***********************************************************************
*** This program compare 2 numbers and tell you which one is larger ***
***********************************************************************

Enter the first number (a): 5
Enter the second number (b): 5
a = 5, b = 5
a == b

Case 2: a > b

***********************************************************************
*** This program compare 2 numbers and tell you which one is larger ***
***********************************************************************

Enter the first number (a): 10
Enter the second number (b): 5
a = 10, b = 5
a > b

Case 3: a < b

***********************************************************************
*** This program compare 2 numbers and tell you which one is larger ***
***********************************************************************

Enter the first number (a): 5
Enter the second number (b): 10
a = 5, b = 10
a < b

The condition check for all 3 cases work as expected.

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000