Task

Say the user is entering numbers, with the last one being 0. In other words, users indicate they are finished by entering 0. The program then displays the sum of all the entered numbers (see Figure 24-1).

Figure 24-1 Summing all numbers until 0

Solution

The solution’s core is accumulating the intermediate result. You have to keep it in a variable and to add every entered number into that variable. As soon as the user terminates the input, the variable will contain the overall sum of all the entered values.

Here’s the code:

static void Main(string[] args)
{
    // Preparations
    int sum = 0;
    int number;

    // Entering numbers until zero
    do
    {
        // Input
        Console.Write("Enter a number (0 = end): ");
        string input = Console.ReadLine();
        number = Convert.ToInt32(input);

        // Adding to intermediate sum
        sum += number;

    } while (number != 0);

    // Output
    Console.WriteLine("Sum of entered numbers is: " + sum.ToString());

    // Waiting for Enter
    Console.ReadLine();
}

Task

In this program the user enters numbers, with the last one being 0 (see Figure 24-2). The program then displays the product of all the entered numbers, with the exclusion of the final 0, obviously, which would make everything 0.

Figure 24-2 Multiplying all numbers

Solution

Here’s the code:

static void Main(string[] args)
{
    // Preparations
    double product = 1;
    int number;

    // Entering numbers until zero
    do
    {
        // Input
        Console.Write("Enter a number (0 = end): ");
        string input = Console.ReadLine();
        number = Convert.ToInt32(input);

        // Accumulating in intermediate product (but not the last zero!)
        if (number != 0)
        {
            product *= number;
        }

    } while (number != 0);

    // Output
    Console.WriteLine("Product of entered numbers (excluding zero) is: " + product.ToString("N0"));

    // Waiting for Enter
    Console.ReadLine();
}

Discussion

Note the following:

• The product variable starts at a value of 1 contrary to 0, which you used when calculating the sum.

• When updating the product, you need to take care not to include the final 0.

• You declare the product variable in type double for the result not to overflow. When you multiply, you quickly get big numbers.

Task

In this program, the user enters ten numbers. The program then outputs which one is the greatest (see Figure 24-3).

Figure 24-3 Outputting the greatest number

Solution

You are going to accumulate the intermediate result again. This time it will be the greatest number “so far.” You have to take special care with the first value; the greatest variable is set to the least possible value at the beginning in order to ascertain that the first entered number is always greater.

Because you expect exactly ten values in the input, it is more convenient to use the for loop here.

Here’s the code:

static void Main(string[] args)
{
    // Preparation
    int greatest = int.MinValue;

    // Input of ten numbers
    for (int order = 1; order <= 10; order++)
    {
        // Input
        Console.Write("Enter " + order.ToString() + ". number: ");
        string input = Console.ReadLine();
        int number = Convert.ToInt32(input);

        // Is it greater than the greatest so far?
        if (number > greatest)
        {
            greatest = number;
        }
    }

    // Output
    Console.WriteLine("The greatest of entered numbers was: " + greatest.ToString());

    // Waiting for Enter
    Console.ReadLine();
}

Task

The task is to choose the second greatest number out of the ten entered ones (see Figure 24-4).

Figure 24-4 Displaying the second greatest number

Solution

You need to remember and continuously update the two greatest numbers greatest . It would not be enough to remember just the second greatest.

The situation resembles a ski competition, with the competitors arriving to the finish line one after another. At a certain moment, someone is the first. Afterward, someone else pushes the first out to the second place. Possibly in a later time, that skier may lose even the second place because someone else will be better than him, or even better than the new leader.

Here’s the code:

static void Main(string[] args)
{
    // Preparation
    int greatest = int.MinValue;
    int secondGreatest = int.MinValue;

    // Input of ten numbers
    for (int order = 1; order <= 10; order++)
    {
        // Input
        Console.Write("Enter " + order.ToString() + ". number: ");
        string input = Console.ReadLine();
        int number = Convert.ToInt32(input);

        // Is it greater than the greatest so far?
        if (number > greatest)
        {
            // Moving so far greatest to the second place
            secondGreatest = greatest;

            // Entered number becomes the greatest so far
            greatest = number;
        }
        else
        {
            // We did not beat the greatest, will we beat the second greatest at least?
            if (number > secondGreatest)
            {
                secondGreatest = number;
            }
        }
    }

    // Output
    Console.WriteLine("The second greatest of entered numbers was: " + secondGreatest.ToString());

    // Waiting for Enter
    Console.ReadLine();
}

Task

You will write a program that repeatedly reads the names entered by the user. The empty input signals the termination. The program then repeats all the entered names, first in the same order and then in the reversed order

Solution

So that you are able to repeat all the names at the end, you need to remember them somewhere. You need to accumulate them. One variable will accumulate them at its end (the same order output) and the other one at its beginning (the reversed order output).

Here’s the code:

static void Main(string[] args)
{
    // Preparation
    string inSameOrder = "";
    string inReversedOrder = "";
    bool terminating;

    // Repeating until blank input
    do
    {
        // Input
        Console.Write("Enter person: ");
        string person = Console.ReadLine();

        // Processing input
        terminating = person.Trim() == "";
        if (!terminating)
        {
            inSameOrder = inSameOrder + person + ", ";
            inReversedOrder = person + ", " + inReversedOrder;
        }

    } while (!terminating);

    // Removing trailing comma and space
    if (inSameOrder.EndsWith(", "))
    {
        int numberOfCharacters = inSameOrder.Length;
        inSameOrder = inSameOrder.Remove(numberOfCharacters - 2);
    }
    if (inReversedOrder.EndsWith(", "))
    {
        int numberOfCharacters = inReversedOrder.Length;
        inReversedOrder = inReversedOrder.Remove(numberOfCharacters - 2);
    }

    // Output
    Console.WriteLine("Entered persons: " + inSameOrder);
    Console.WriteLine("In reversed order: " + inReversedOrder);

    // Waiting for Enter
    Console.ReadLine();
}

Discussion

Note the following:

• You use the Trim method to cut off possible leading and trailing spaces of the entered text in order to allow termination with any blank input, including several spaces.

• At the end, you have to get rid of the last two characters in both accumulated pieces of text. Before doing it, you test whether these two characters (a comma and a space) appear at the end of the text at all. These characters will not be there if the user immediately terminates the program by entering a blank line.

• To test whether text ends with something, you use the EndsWith method.