Scala_for_impatients_Ch2

Note#

An if expression has a value.
A block has a value—the value of its last expression.
The Scala for loop is like an “enhanced” Java for loop.
- for ( i -> 0 to n)
- for ( i -> 0 until n)
Semicolons are (mostly) optional.
- Need when: more than one statements on a single line
The void type is Unit.
Avoid using return in a function.
- With a recursive function, you must specify the return type.
Beware of missing = in a function definition.
- Because the procedure doesn’t return any value, we omit the = symbol
- It is a common error to accidentally omit the = in a function definition.You then get an error message at the point where the function is called, and you are told that Unit is not acceptable at that location.
Exceptions work just like in Java or C++, but you use a “pattern matching” syntax for catch.
Scala has no checked exceptions.
Variable Arguments
- The remedy is to tell the compiler that you want the parameter to be considered an argument sequence. Append : _*, like this:
  - val s = sum(1 to 5: _*) // Consider 1 to 5 as an argument sequence

Exercises#

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.

Answer:
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def signum(i:Int)={
if (i>0) 1 else {
if (i<0) -1 else 0
}
}
What is the value of an empty block expression {}? What is its type?
Answer: Unit
Come up with one situation where the assignment x = y = 1 is valid in Scala.
(Hint: Pick a suitable type for x.)

Answer: x is type of Unit
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var y = 1
var x = {}
x = y = 1

Write a Scala equivalent for the Java loop
for (int i = 10; i >= 0; i–) System.out.println(i);

Answer:

for(i<- 0 to 10) println(10-i)

// another version of Answer

for(i <- 0 to 10 reverse)print(i)

Write a procedure countdown(n: Int) that prints the numbers from n to 0.

Answer:

def countdown(n:Int) {
  for(i<- 0 to n reverse) print(i)
}

// another version:  

def countdown(n:Int) {
  0 to n reverse foreach print
}

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 825152896.

Answer:
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def productUcode(str: String):Long = {
  var result: Long = 1
  for(c<-str) result *= c.toLong
  result
}

Solve the preceding exercise without writing a loop. (Hint: Look at the StringOps Scaladoc.)

Answer:
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def productUcode(str: String):Long = {
  var result: Long = 1
  str.foreach(result *= _.toLong)
  result
}

Write a function product(s : String) that computes the product, as described in the preceding exercises.

same with above

Make the function of the preceding exercise a recursive function

Answer:

def productUcode2(str:String):Long = {
  if(str.length == 1) return str.head.toLong
  else str.head.toLong * productUcode2(str.tail)
}

Write a function that computes xn, where n is an integer. Use the following recursive definition:
• xn = y2 if n is even and positive, where y = xn / 2.
• xn = x· xn – 1 if n is odd and positive.
• x0 = 1.
• xn = 1 / x–n if n is negative.
Don’t use a return statement.

Answer:
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def computesXn(x:Double, n:Int):Double = {
  if (n==0) 1.toDouble else {
    if (n < 0) 1/computesXn(x, -n) else {
      if (n%2 == 0) {
        val y = computesXn(x, n/2); y*y
        } else {
        x*computesXn(x, n-1)
      }
    }
  }
}

//

def computesXn(x:Double, n:Int):Double = {
  if (n==0) 1.toDouble
  else if (n>0 && n%2==0) {
    val y = computesXn(x, n/2); y*y
  }
  else if (n>0 && n%2==1) {
    x*computesXn(x, n-1)
  }
  else 1/computesXn(x, -n)
}

Note#

Exercises#

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