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2. Loops and Conditionals in Java

Start by making a new Eclipse project named task02. Give it a package named ua.khpi.oop.your_first_name.task02.

1) Add a class called LoopEx to your package. Give it a method named printNums1 that will take a number and print out the numbers from 0 up to and including that number. For example, if you pass in 3, it should print 0, 1, 2, and 3. Your overall code will look roughly like this:

package your_package_name;

public class LoopEx {
	public static void main(String[] args) {
		printNums1(3);
	}
	public static void printNums1(int upperLimit) {
		// TODO: Print out the numbers from 0 up to and including upperLimit.
	}
}

2) Add a second method named printNums2 to your class. This one should print every other number: i.e., 0, 2, 4, etc., up to the last value that is less than or equal to the argument to the method. For example, if your main method calls printNums2(7), it should print 0, 2, 4, and 6. Remember that i++ in a loop is the same as if you had written i = i + 1.

3) Add a third method named printNums3 to your class. This one should be similar to your first method, but should print in reverse order. For example, if your main method calls printNums3(5), it should print 5, 4, 3, 2, and 1.

4) Copy the ArrayEx class into your project.

public class ArrayEx {
	public static void main(String[] args) {
		double[] numbers = { 1.1, 2.2, 3.3 };
		System.out.print("Array: ");
		printArray(numbers);
		System.out.println();
		System.out.println("calcSum1(): " + calcSum1(numbers));
		System.out.println("calcSum2(): " + calcSum2(numbers));
		System.out.println("calcSum3(): " + calcSum3(numbers));
		System.out.println("calcSum4(): " + calcSum4(numbers));
	}

	public static void printArray(double[] numbers) {
		// TODO: Print each element of the array as follows: {numbers[0], numbers[1], ...}.
	}

	public static double calcSum1(double[] numbers) {
		double sum = 0;
		for (double num : numbers) {
			sum = sum + num; // Or sum += num
		}
		return sum;
	}

	public static double calcSum2(double[] numbers) {
		double sum = 0;
		for (int i = 0; i < numbers.length; i++) {
			sum = sum + numbers[i];
		}
		return sum;
	}

	public static double calcSum3(double[] numbers) {
		double sum = 0;
		int i = 0;
		while (i < numbers.length) {
			sum = sum + numbers[i];
			i++; // Or i = i + 1, or i += 1
		}
		return sum;
	}

	// Unlike the other three versions, this one fails for a 0-length array.
	public static double calcSum4(double[] numbers) {
		double sum = 0;
		int i = 0;
		do {
			sum = sum + numbers[i];
			i++;
		} while (i < numbers.length);
		return sum;
	}
}

5) Implement the printArray method to print an array to the screen. You should obtain the following result:

Array: {1.1, 2.2, 3.3}
calcSum1(): 6.6
calcSum2(): 6.6
calcSum3(): 6.6
calcSum4(): 6.6

6) Add a method named calcAverage that, given an array of doubles, will return the average value.

public static double calcAverage(double[] numbers) {
	// TODO: Return the average value of 'numbers'.
	return 0.0;
}

Hint: Have your new method make use of the existing calcSum1 method, and then use the length property of arrays.

For example, given the array already made inside main (containing 1.1, 2.2, and 3.3), calling calcAverage(numbers) will return 2.1999... (almost 2.2, but not exactly, due to roundoff error).

7) Add a method named numPositive that, given an array of doubles, will return the count (int) of how many of them are greater than or equal to zero.

public static int numPositive(double[] numbers) {
	// TODO: Return the count of how many of the array entries are greater than or
	// equal to zero.
	return 0;
}

For example, given the array already made inside main, calling numPositive(numbers) will output 3. Add a few negative numbers to the array and verify that you still get 3. Add a new positive number and verify that you now get 4.

8) Add a method named numInRange that, given an array of numbers, a lower bound, and an upper bound, will return the count of how many of the array entries are between the two bounds, inclusive.

public static int numInRange(double[] nums, double lowerBound, double upperBound) {
	// TODO: Return the count of how many of the array entries are between the two
	// bounds, inclusive.
	return 0;
}

For example, given the array already made inside main (containing 1.1, 2.2, and 3.3), calling numInRange(numbers, 1.1, 3.1) should return 2.

9) Add the following code to main() to test all implemented methods:

System.out.println("Average of {1.1,2.2,3.3} = " + calcAverage(numbers));
System.out.println("Number positive = " + numPositive(numbers));
double[] numbersWithNegative = { 1.1, 2.2, 3.3, -1, -2};
System.out.print("Array: ");
printArray(numbersWithNegative);
System.out.println();
System.out.println("Number positive = " + numPositive(numbersWithNegative));
double[] moreNumbers = { 1.1, 2.2, 3.3, -1, -2, 4 };
System.out.print("Array: ");
printArray(moreNumbers);
System.out.println();
System.out.println("Number positive = " + numPositive(moreNumbers));
System.out.print("Array: ");
printArray(numbers);
System.out.println();
System.out.println("Number from 1.1 to 3.2 = " + numInRange(numbers, 1.0, 3.1));

You should obtain the following result:

Average of {1.1,2.2,3.3} = 2.1999999999999997
Number positive = 3
Array: {1.1, 2.2, 3.3, -1.0, -2.0}
Number positive = 3
Array: {1.1, 2.2, 3.3, -1.0, -2.0, 4.0}
Number positive = 4
Array: {1.1, 2.2, 3.3}
Number from 1.1 to 3.2 = 2


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