CS计算机代考程序代写 c# algorithm KIT506 Programming Practical 2: Arithmetic, Program Tracing & Decisions

KIT506 Programming Practical 2: Arithmetic, Program Tracing & Decisions

KIT506 Programming Practical 2: Arithmetic, Program Tracing & Decisions

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Aims:

to gain experience with the arithmetic operators; and
to implement a simple algorithm that utilises variables of C#’s value types and strings

1 Reading and Tracing Exercises

1.1 Reading arithmetic

Suppose that the following statements were placed in the Main method of a new C# program. What would happen when the program is executed? Answer the questions below.

1
2
3
4
5
6

int a, b, c;

a = 10;
b = 8;
c = a + b;
a = c / b;

What value does each variable have after each of these lines has been executed? (Create a table like the one below, or take a copy of the code and write down the values next to each line.)

Line a b c
3 10 no value yet no value yet
4 10 8 no value yet
5 10 8 18
6 2 8 18

1.2 Constants, assignment and arithmetic

Suppose the following statements were placed in the Main method of a different C# program. What would happen as the program is executed?

1
2
3
4
5
6
7

const int PRIME = 31;
const int DIVISOR = 3;
int a, b;

a = PRIME;
b = 2;
a = (a + b) / DIVISOR;

What value does each variable after each of the indicated lines has been executed?

Line PRIME DIVISOR a b
1 31
2 31 3
5 31 3 31
6 31 3 31 2
7 31 3 11 2

After lines 1 and 2, can the values of PRIME and DIVISOR ever change? No, because they are constants, declared with the keyword const.

Tip: When reading someone else’s code for the first time, consider creating a tracing table to help you follow what’s going on.

1.3 What’s the output?

Suppose the following lines of code were placed in the Main method of a C# program (and were the only contents of the Main method). What would the output of the program be?

int height = 176; //height in cm
double mass = 67.4; //mass in kg
double bmi;

bmi = mass/Math.Pow(height/100.0, 2);

Console.Write(“Calculating BMI using height of ” + height + ” cm “);
Console.WriteLine(“and weight of ” + mass + ” kg.”);
Console.WriteLine(“BMI: ” + bmi);

First, this is not the cleanest or most useful implementation of a BMI calculator. It could be improved by, at least:

requesting the values of height and weight (mass) from the user; and
storing height as a double representing the height in metres.

But, what’s the output?

Calculating BMI using height of 176 cm and weight of 67.4 kg.
BMI: 21.7587809917355

Let’s break it down:

The value of bmi is calculated according to mass/Math.Pow(height/100.0, 2), which is evaluated in the following order:

height/100.0, which performs real-valued division on height (100.0 is a double) to produce the double value 1.76.
Math.Pow(1.76, 2), which is the equivalent to 1.762 (Math.Pow(x, y) calculates xy), producing the value 3.0976.
mass/3.0976, which evaluates to 21.76.

And did you expect two or three lines to be printed?

Because the first part of the output is produced by Console.Write(), it doesn’t add a newline character at the end and its output appears on the same line as the next Console.WriteLine() statement’s.

2 Newspaper Article Writer

The Console.ReadLine() method allows your console application to read a line of text entered by the user. Create a new program with the following statements in the Main() method:

string name;

Console.Write(“Enter your name: “);
name = Console.ReadLine();

Console.WriteLine(“Name is ” + name);

That’s quite boring, so modify your program as follows:

Declare two more variables, int age and string country.
Add more user prompts and Console.ReadLine()s to read the user’s age and country into those variables. When reading the int you will need to use Int32.Parse(Console.ReadLine()), which will convert the text the user entered (that looks like a number but is just text) into the internal binary representation of an integer value.
Modify the message at the end to be similar to the following, where each italicised value comes from the variable with that name:

name (age), of country, was this year’s surprise Eurovision Song Contest winner. This is country’s first win in the contest.

string name;
string country;
int age;

Console.Write(“Enter your name: “);
name = Console.ReadLine();
Console.Write(“Enter your age: “);
age = Int32.Parse(Console.ReadLine());
Console.Write(“Enter your country of origin: “);
country = Console.ReadLine();

Console.WriteLine(name + ” (” + age + “), of ” + country + “, was this year’s surprise Eurovision Song Contest winner. This is ” + country + “‘s first win in the contest.”);

There’s another way you can format strings containing variable values in C#, which may prove useful later (it works with labels on graphical user interfaces, too). You can write a format string as the first argument to WriteLine() and then place all the variables after it, separated by commas. The basic form of the format string is to refer to each variable you want to place in the output text by its position (which starts from 0):

Console.WriteLine(“{0} ({1}), of {2}, was this year’s surprise Eurovision Song Contest winner. This is {2}’s first win in the contest.”, name, age, country);

Tip: In addition to Int32.Parse(Console.ReadLine()) for reading integer values from the console, you can read real-valued numbers (type double) with Double.Parse(Console.ReadLine()) and bool values (true or false) with Boolean.Parse(Console.ReadLine()).

3 Making some simple decisions

Create a new program that prompts the user for a positive integer n and stores it in a variable. Then apply the following rules:

if n is even your program should divide it by 2 and store the result back in n (that is, overwrite the value of n);
otherwise it should multiple n by 3 and add 1, and store the result back in n

And then display the new value of n. Hint: you can determine if a number is even by checking if the remainder from dividing it by 2 is equal to zero (n % 2 == 0 in C#).

int n;

Console.Write(“Enter a positive integer: “);
n = Int32.Parse(Console.ReadLine());

if (n % 2 == 0) //n is even
{
n = n / 2;
}
else
{
n = 3 * n + 1;
}

Console.WriteLine(“The final value of n is ” + n);