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.

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

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

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

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

Exercise 2-7

Write a program that count down from 10 to -5

Solution

Though it is tempting to just create a for loop (or a while loop) with an invariant that start from 10, output the invariant, increment the invariant by -1, and exit the loop when the invariant becomes smaller than -5, I have decided to write the program in a way that is consistent with the asymmetric range method that the author has taught us. i.e. make use of the asymmetric invariant range denotation i[0,End) – meaning creating an invariant i that start with the value of zero (and always zero!), with an end number representing the number of loops that we wish the invariant to go through.

So how do we find out how many loops we need?

• If we are to only output 10 we need 1 loop (to display 1 number). 10 – 10 + 1 = 1. i.e. we need the asymmetric range of [0,1)
• If we need to output 10 and 9, we need 2 loops (to display 2 numbers). 10 – 9 +1 = 2. i.e. we need the asymmetric range of [0,2)
• If we need to output 10, 9, 8, we need 3 loops (to display 3 number). 10 – 8 + 1 = 3. i.e. we need the asymmetric range of [0,3)

See the pattern?

If we need to display N numbers (counting down), we need N loops, as calculated as:

N = (( EndNumber - StartNumber ) * (-1 ) ) + 1

So if we are to display 10, 9, 8 …. -3, -4, -5, we have:

• StartNumber = 10
• EndNumber = -5
• NumLoops = (( EndNumber – StartNumber ) * (-1 ) ) + 1 = ( (-5 -10) * (-1) ) + 1 = 15 + 1 = 16 loops.

Our asymmetric invariant range therefore takes this denotation: i [0, 16)

Understanding these relationships will allow us to be consistent in the way we program. This knowledge will enable us to be robust in solving a problem like this: write a program that count down from 999 to – 777. i.e. we can simply repeat the above methodology (or algorithm) without scratching our heads and get stuck!

One more question before we move on, what happen if we are counting up instead of counting down?

e.g. if we are to display -5, -4, -3, … 8, 9, 10, we have:

• StartNumber = -5
• EndNumber = 10
• NumLoops = ( ( EndNumber – StartNumber ) * 1 ) + 1 = ( ( 10 – (-5) ) * 1 ) + 1 = 15 + 1 =16 loops.

If we need to display N numbers (counting up), we need N loops, as calculated as:

N = (( EndNumber - StartNumber ) * (+1 ) ) + 1

Our asymmetric invariant range still take the same denotation: i [0, 16)

The general equation for NumLoops is therefore:

NumLoops = ( (EndNumber - StartNumber) * directionFlag ) + 1

• If counting up (i.e. EndNumber > StartNumber), the directionFlag is 1.
• Otherwise, the directionFlag is -1.

Quite initiative really as we have a positive flag for counting up, and a negative flag for counting down.

Also note that if EndNumber == StartNumber, directionFlag becomes irrelevant because in that case the numLoops will be (0 * directionFlag) + 1, which is always 1.

The directionFlag also provides us a mean to determine whether we wish to increment the output value by +1 or -1.

The last bit is the actual counting and displaying the numbers. This is how I’d do it. (Note: there are many ways to do this!)

// Do the counting and display result row by row
for (int i = 0, iOut = startNumber; i != numLoops; ++i)
{
    std::cout << "Loop " << i << "| iOut = " << iOut << std::endl;
    iOut += directionFlag;
}

This is a for loop with i [0,numLoops) denotation. i.e. we are certain that there will be numLoops number of loops in total.

The first loop output the start value. At the end of the loop we increment the invariant by either +1 (counting up), or -1 (counting down). If the condition (i != numLoops) is true, loop again. Otherwise, the loop terminates.

The above snippet essentially isolate the invariant i component from the computed iOut component – for flexibility. In other words, the function of the invariant is to purely count the number of completed loops. We leave the computed component as free as it can be.

Putting everything together, we now have our full program:

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

int main()
{
    // Display greeting message
    cout << "*****************************n"
         << "*** The Counting Program ****n"
         << "*****************************n";

    // Obtain startNumber and endNumber from the user
    cout << endl;
    cout << " Enter Start Number: ";
    int inputStartNumber;
    cin  >> inputStartNumber;
    cout << " Enter End Number: ";
    int inputEndNumber;
    cin  >> inputEndNumber;

    const int startNumber = inputStartNumber;
    const int endNumber = inputEndNumber;

    // Compute the directionFlag and numLoops
    int directionFlag, numLoops;
    if (endNumber > startNumber) {directionFlag = 1;}
    else {directionFlag = -1;}

    numLoops = ( (endNumber - startNumber) * directionFlag ) + 1;

    // Display the directionFlag and numLoops
    cout << " directionFlag = " << directionFlag
         << " | numLoops = " << numLoops << endl;

    // Do the counting and display result row by row
    for (int i = 0, iOut = startNumber; i != numLoops; ++i)
    {
        std::cout << "Loop " << i << "| iOut = " << iOut << std::endl;
        iOut += directionFlag;
    }
    return 0;
}

Result

The code compile okay. Let’s run it and see what we get!

Case 1: count from 10 to -5

*****************************
*** The Counting Program ****
*****************************

 Enter Start Number: 10
 Enter End Number: -5
 directionFlag = -1 | numLoops = 16
Loop 0| iOut = 10
Loop 1| iOut = 9
Loop 2| iOut = 8
Loop 3| iOut = 7
Loop 4| iOut = 6
Loop 5| iOut = 5
Loop 6| iOut = 4
Loop 7| iOut = 3
Loop 8| iOut = 2
Loop 9| iOut = 1
Loop 10| iOut = 0
Loop 11| iOut = -1
Loop 12| iOut = -2
Loop 13| iOut = -3
Loop 14| iOut = -4
Loop 15| iOut = -5

Case 2: count from -5 to 10

*****************************
*** The Counting Program ****
*****************************

 Enter Start Number: -5
 Enter End Number: 10
 directionFlag = 1 | numLoops = 16
Loop 0| iOut = -5
Loop 1| iOut = -4
Loop 2| iOut = -3
Loop 3| iOut = -2
Loop 4| iOut = -1
Loop 5| iOut = 0
Loop 6| iOut = 1
Loop 7| iOut = 2
Loop 8| iOut = 3
Loop 9| iOut = 4
Loop 10| iOut = 5
Loop 11| iOut = 6
Loop 12| iOut = 7
Loop 13| iOut = 8
Loop 14| iOut = 9
Loop 15| iOut = 10

Case 3: count from 10 to 10

*****************************
*** The Counting Program ****
*****************************

 Enter Start Number: 10
 Enter End Number: 10
 directionFlag = -1 | numLoops = 1
Loop 0| iOut = 10

 Reference

Exercise 2-6

What does the following code do?

int i = 0;
while (i < 3)
{
    i += 1;
    std::cout << i <<std::endl;
}

Solution

This is very a classic usage of while loop with the following properties:

Invariant i, with the initial integer value of zero.

int i = 0;

And asymmetric range [0,3). i.e. invariant i is incremented at a specified interval (in this case 1) from the initial value of 0 to the ending value of 3 (but excluding 3). Note the mixed use of square and round brackets (in the sense of writing in the post rather than the actual code.). We use square brackets [ ] to imply inclusiveness, and round bracket ( ) to imply exclusiveness.

while (i < 3)    // asymmetric range of [0,3)
{
    i += 1;     // invariant increment interval is 1
    std::cout << i <<std::endl;   // display the invariant to the standard output console
}

I denote this while loop as i [0,3)

Using logic, I’d expect the code to do this:

To confirm this let’s run the following code and output result.

#include <iostream>

int main()
{
     int i = 0;
     while (i < 3)
     {
         i += 1;
         std::cout << i <<std::endl;
         //i += 1;
     }
     return 0;
}

Result

1
2
3

The nice thing about using asymmetric range [0,N) is the ease of understanding. i.e. The invariant is true for N number of loops.

Extras

Say now if we modify the code so that the invariant increment happens at the bottom. i.e.

int i = 0;
while (i < 3) {
    std::cout << i <<std::endl;
    i += 1;
}

Our result should now look like this:

We run the modified code and it confirms our expectation:

0
1
2

Reference

Exercise 2-4

The framing program (Exercise 2-0) writes the mostly blank lines that separate the borders from the greeting one character at a time. Change the program so that it writes all the spaces needed in a single output expression.

Solution

The key to solving this problem is to use the string constructor statement that we learnt from chapter 1 on string.

std::string z(n, c) ;

This statements defines z as a variable of types std::string that initially contains n copies of the character c. Here, c must be a char, not a string or a string literal.

For example, the statement std::string z(10, ‘ ‘); will create a string variable z that is occupied of 10 space characters (char). We define char values using single quotes. (double quotes are for defining string values).

Bearing this in mind, I now demonstrate the high level Solution Strategy.

The program will need to do the followings:

• Ask the user for his first name so we can build the constant string greeting.
• Ask the user for the padding value so we can (in conjunction with the readily build greeting string) compute the total frame rows and columns. We will also be able to build the constant string padStringLeftRight and padStringTopBottom.
• Create an asymmetric for loop 1 r[0,rows) – the invariant r tells us which row we are at.
• At the beginning of each row, reset the invariant c to 0 – i.e. always start from column 0 (the first column of the row).
• Create an asymmetric while loop 2 c[0,cols) within loop 1 – the invariant c tells us which column we are at, keeping the row constant.
• We perform the following checks during each complete loop to determine what to output:

1. Are we at the exact position to output the greeting string? If so, output the constant string greeting, and increment the invariant c by size of the string. Else…
2. Are we on any border? if so, output a string literal of 1 asterisk character, and increment the invariant c by 1. Else…
3. Are we on the greeting line? if so, output the constant string padStringLeftRight, and increment the invariant c by size of the string. Else…
4. Output the constant string padStringTopBottom, and increment the invariant c by size of the string.

Notes:

• We use a for loop 1 (instead of while loop) because the increment of the invariant r (for rows) s always by 1 – using a for loop is easier and clearer. Nevertheless if we want it is also okay to use a while loop – just a matter of preference really.
• We use a while loop 2 (instead of a for loop) because the increment of the invariant c (for columns) changes depending on what we output. Sometimes the increment is 1 (asterisk), sometimes the size of the string (greeting, padStringLeftRight, padStringTopBottom).

Putting this all together, we now have our complete program.

#include <iostream>
#include <string>

// say what standard-library names we use
using std::cin;
using std::endl;
using std::cout;
using std::string;

int main()
{
    // ask for the person's name
    cout << "Please enter your first name: ";

    // read the name
    string name = "";
    cin >> name;

    // ask user to provide the padding between frame border and greeting message
    cout << "Please specify an integer padding value: ";
    int inPad = 0;
    cin >> inPad;

    // build the message that we intend to write
    const string greeting = "Hello, " + name + "!";

    // build the constant padding parameters
    const int pad = inPad; // padding

    // compute the number of rows to write
    const int rows = ( pad * 2 ) + 3;

    // compute the number of columns to write
    const string::size_type cols = greeting.size() + ( pad * 2 ) + 2;

    // build the padding string that lives on the same row as the greeting, between the greeting and the border.
    const string padStringLeftRight( pad, ' ' );

    // build the padding string that lives between the top (or bottom) border and and greeting.
    const string padStringTopBottom( ( cols - 2 ) , ' ' );

    // write a blank line to separate the output from the input
    cout << endl;

    // write 'rows' rows of output
    // invariant: we have written r rows so far
    for ( int r = 0; r != rows ; ++r )
    {
        string::size_type c = 0;

        // invariant: we have written c characters so far in the current row
        while ( c != cols)
        {
            // Are we at the exact position where the greeting should start?
            if ( ( r == pad + 1 ) && ( c == pad + 1 ) )
            {
                // output the greeting message
                cout << greeting;
                c += greeting.size();
            }
            else
            {
                // are we on the border?
                if (
                       ( r == 0 )         // top row
                    || ( r == rows - 1 )  // or bottom row
                    || ( c == 0 )         // or left-most column
                    || ( c == cols - 1 )  // or right-most column
                   )
                {
                    // output the border asterisks
                    cout <<  "*";
                    ++c;
                }
                else
                {
                    // are we on the greeting line?
                    if ( r == pad + 1)
                    {
                        // output the empty spaces between left (or right) border and the greeting
                        cout << padStringLeftRight;
                        c += padStringLeftRight.size();
                    }

                    else
                    {
                        // output the empty spaces between top (or bottom) border and the greeting
                        cout << padStringTopBottom;
                        c += padStringTopBottom.size();
                    }
                }
            }
        }
        cout << endl;
    }
    return 0;
}

Further explanations

Just little bit more explanation, just for peace of mind.

To compute the total rows required by the frame, we sum the:

• top and bottom padding (pad * 2),
• top border, bottom border, and the greeting lines (1 + 1 + 1 = 3).

Hence we have this line in the code:

    // compute the number of rows to write
    const int rows = ( pad * 2 ) + 3;

To compute the total columns required by the frame, we sum the:

• The size of the greeting message ( greeting.size() )
• The size of the left and right paddings combined ( pad * 2 )
• The size of left and right borders ( 1 + 1 = 2 ),

Also, as per the book chapter 2.4 (page 22), it is good habit to use the type string::size_type (instead of integer) when computing the sum of string sizes.

Hence we have this line in the code:

    // compute the number of columns to write
    const string::size_type cols = greeting.size() + ( pad * 2 ) + 2;

Result

The program compiles okay. Here is our output.

Please enter your first name: Johnny
Please specify an integer padding value: 3

**********************
*                    *
*                    *
*                    *
*   Hello, Johnny!   *
*                    *
*                    *
*                    *
**********************

Reference

Exercise 2-3

Rewrite the framing program (Exercise 2-0) to ask the user to supply the amount of spacing to leave between the frame and the greeting.

Solution

Turns out that in my solution to Exercise 2-2 I already have covered this. i.e. to ask user to supply the horizontal and vertical padding separately. Please see my solution to Exercise 2-2 for details.

Reference

Exercise 2-2

Change the framing program (Exercise 2-0) so that it uses a different amount of space to separate the sides from the greeting than it uses to separate the top and bottom borders from the greeting.

Solution

The original framing program in Exercise 2-0 “hard-codes” the const int pad which forces both vertical and horizontal paddings to be identical. The question asks for a capability to assign the vertical padding and horizontal padding separately.

To do this I am going to adjust the original framing program taking into account of the followings:

• Ask the user to provide the vertical padding, and store it as an int variable inPadY.
• Ask the user to provide the horizontal padding, and store it as an int variable inPadX.
• Assign the int variable inPadY to a constant integer padY.
• Assign the int variable inPadX to a constant integer padX.
• Replace all the (old) pad references associating with rows with padY (e.g. the constant rows and invariant r). Because the number of rows depends on vertical padding.
• Replace all the (old) pad references associating with columns with padX (e.g. the constant cols and invariant c). Because the number of columns depends on horizontal padding.

The modified program looks like this:

#include <iostream>
#include <string>

// say what standard-library names we use
using std::cin;
using std::endl;
using std::cout;
using std::string;

int main()
{
    // ask for the person's name
    cout << "Please enter your first name: ";

    // read the name
    string name;
    cin >> name;

    // ask user to provide the vertical padding
    cout << "Please provide integer padY (vertical padding): ";
    int inPadY;
    cin >> inPadY;

    // ask user to provide the horizontal padding
    cout << "Please provide integer padX (horizontal padding): ";
    int inPadX;
    cin >> inPadX;

    // build the message that we intend to write
    const string greeting = "Hello, " + name + "!";

    // build the constant padding parameters
    const int padY = inPadY; // vertical padding
    const int padX = inPadX; // horizontal padding

    // the number of rows to write
    const int rows = ( padY * 2 ) + 3;

    // the number of columns to write
    const string::size_type cols = greeting.size() + ( padX * 2 ) + 2;

    // write a blank line to separate the output from the input
    cout << endl;

    // write 'rows' rows of output
    // invariant: we have written r rows so far
    for ( int r = 0; r != rows ; ++r )
    {
        string::size_type c = 0;

        // invariant: we have written c characters so far in the current row
        while ( c != cols)
        {
            // is it time to write the greeting?
            if ( ( r == padY+ 1 ) && ( c == padX + 1 ) )
            {
                cout << greeting;
                c += greeting.size();
            }
            else
            {
                // are we on the border?
                if (
                       ( r == 0 )         // top row
                    || ( r == rows - 1 )  // or bottom row
                    || ( c == 0 )         // or left-most column
                    || ( c == cols - 1 )  // or right-most column
                   )
                {
                    cout <<  "*";
                }
                else
                {
                    cout << " ";
                }
                ++c;
            }
        }
        cout << endl;
    }
    return 0;
}

Result

Case 1: padY = 1; padX = 10

Please enter your first name: Johnny
Please provide integer padY (vertical padding): 1
Please provide integer padX (horizontal padding): 10

************************************
*                                  *
*          Hello, Johnny!          *
*                                  *
************************************

Case 2: padY = 5; pad X = 5

Please enter your first name: Johnny
Please provide integer padY (vertical padding): 1
Please provide integer padX (horizontal padding): 10

************************************
*                                  *
*          Hello, Johnny!          *
*                                  *
************************************

Case 3: padY = 10 padX = 1

Please enter your first name: Johnny
Please provide integer padY (vertical padding): 5
Please provide integer padX (horizontal padding): 5

**************************
*                        *
*                        *
*                        *
*                        *
*                        *
*     Hello, Johnny!     *
*                        *
*                        *
*                        *
*                        *
*                        *
**************************

So far, so good!

Reference