scala for impatient exercise chapter 2

ex2.scala

/*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(x: Int): Int = if(x>0) 1 else -1

/*
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)

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

def countDown(n: Int): Unit= if(n == 0) println(0) else if(n>0) {println(n);countDown(n-1)}

/*
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 825152896.
*/
var prod =0
for(c <- "Hello") prod*=c.toInt

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

*/
var prod =0
"Hello".foreach(prod*=_.toInt)


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

                               
def product(s:String) :Int = { 
 var prod =1                    
 s.foreach(prod*=_.toInt)       
 prod                           
}

/*
9. Make the function of the preceding exercise a recursive function.
*/
                                      
def prodR(s:String):Int = {                  
 if(s.length() == 1) s.head.toInt
 else s.head.toInt * prodR(s.drop(1))                                    
}                                                                   

Prob 6 result has overflown. Do you want to try BigInt?

L32: maybe var prod: BigInt = 1 ?

var prod =0
should be
var prod = 1