Scala for the Impatient: Chapter 2 exercises solutions

I’ve started going through the book: Scala for the Impatient by Cay Horstmann. I’m posting the solution to chapter 2 exercises below.

	// 1. The signum of a number is 1 if the number is positive, -1 if it is negative, and
	// 0 if it is zero. Write a function that computes this value.
	def signum(n: Int) = {
	if (n > 0) 1
	else if (n < 0) -1
	else 0
	} //> signum: (n: Int)Int
	signum(4) //> res0: Int = 1
	signum(0) //> res1: Int = 0
	signum(-20) //> res2: Int = -1
	// 2. What is the value of an empty block expression {}? What is its type?
	val blank = {} //> blank : Unit = ()
	// Value is () and type is Unit
	// 3. Come up with one situation where the assignment x = y = 1 is valid in Scala.
	// (Hint: Pick a suitable type for x.)
	var y: Int = 0 //> y : Int = 0
	val x: Unit = y = 1 //> x : Unit = ()
	// x should be of type Unit
	// 4. Write a Scala equivalent for the Java loop
	// for (int i = 10; i >= 0; i--) System.out.println(i);
	for(i <- 10 to (0, -1)) println(i) //> 10
	//| 9
	//| 8
	//| 7
	//| 6
	//| 5
	//| 4
	//| 3
	//| 2
	//| 1
	//| 0
	// 5. Write a procedure countdown(n: Int) that prints the numbers from n to 0.
	def countdown(n: Int): Unit = {
	for(i <- n to (0, -1)) println(i)
	} //> countdown: (n: Int)Unit
	countdown(5) //> 5
	//| 4
	//| 3
	//| 2
	//| 1
	//| 0
	// 6. Write a for loop for computing the product of the Unicode codes of all letters
	// in a string. For example, the product of the characters in "Hello" is 9415087488.
	var prod: BigInt = 1 //> prod: BigInt = 1
	for(c <- "Hello") prod *= c
	prod //> res3: Int = 9415087488
	// 7. Solve the preceding exercise without writing a loop. (Hint: Look at the StringOps
	// Scaladoc.)
	"Hello".foldLeft(1: BigInt)((a, b) => a * b) //> res4: BigInt = 9415087488
	// 8. Write a function product(s : String) that computes the product, as described
	// in the preceding exercises.
	def product(s: String) = s.foldLeft(1: BigInt)((a, b) => a * b)
	//> product: (s: String)BigInt
	product("Hello") //> res5: BigInt = 9415087488
	// 9. Make the function of the preceding exercise a recursive function.
	def productRec(s: String): BigInt = {
	if (s.length == 0) 1
	else s.head * productRec(s.tail)
	} //> productRec: (s: String)BigInt
	productRec("Hello") //> res6: BigInt = 9415087488
	// 10. Write a function that computes x^n, where n is an integer. Use the following
	// recursive definition
	// xn = y2 if n is even and positive, where y = x ^ n / 2.
	// xn = x * x ^ n - 1 if n is odd and positive.
	// x0 = 1
	// xn = 1 / x ^ -n if n is negative.
	// Don't use a return statement.
	def xpown(x: BigDecimal, n: Int): BigDecimal = {
	if (n == 0) 1
	else if (n < 0) 1 / xpown(x, -n)
	else if (n % 2 == 0) {
	val i = xpown(x, n / 2)
	i * i
	}
	else x * xpown(x, n - 1)
	} //> xpown: (x: BigDecimal, n: Int)BigDecimal
	xpown(2, 1024) //> res7: BigDecimal = 1.797693134862315907729305190789023E+308
	xpown(-2, -10) //> res8: BigDecimal = 0.0009765625
	xpown(-2, 10) //> res9: BigDecimal = 1024

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