CSCI A201/A597 and I210

Lecture Notes Ten

Second semester 2000-2001

Boolean values, expressions, and if statements.
What's this? 
x < 15
 A boolean expression: an expression whose value is either true or false.
Can you define boolean variables in Java? Sure, boolean is a primitive type in Java. 
boolean b;
 b can only hold true and false values.
How do you read this? 
x <= 15
 "x is less than or equal to 15"
How do you write "x is in between 9 and 15"? First, what value does it have?
We don't know yet, it depends on what x is. So we might as well call it p (x).
Very well; now we can look at it for particular values of x. p (3) is false
p (20) also is false Come to think of it, p (x) is false for many values.
What's p (x) again? A statement about x being between 9 and 15.
When is it true? When 
x <= 15
and at the same time 
x >= 9
How do you write AND in Java? &&
So p (x) can be written as: && is read as AND 
(x >= 9) && (x <= 15)
And || is read as OR. While ! stands for NOT in Java.
I like A201 ! I do ! think this is that funny...
So &&, ||, and are operators for truth values. Yes. How do they work?
Let me draw a table.
p q p && q p || q ! q
true true true true false
true false false true true
false true false true
false false false false
AND OR NOT
So && works in the following way: you graduate if you satisfy both of two requirements.
Otherwise you don't graduate. || is a bit more lenient.
You graduate if you satisfy at least one of the two requirements. And only when none of them has been satisfied you do not graduate.
And ! is easy. It is, indeed.
The boolean type is called after mathematician George Boole, a pioneer in the study of logic. Logic is tricky: suppose a is a boolean value, true or false. What value does 
a || !a
Doesn't it depend on the value of a? No.
Well, then let's look at all possible cases:
a ! a a || (! a)
true false true
false true true
Doesn't it look easy now?
Yes, and there weren't even too many cases. How do you compute? 
3 + 5 * 2
Why are you bringing this up? Because as you know there is an implicit order of evaluation for arithmetic expressions.
Does a similar set of rules apply to boolean expressions? Yes. In arithmetic, unary minuses are taken into account before we do any multiplications...
... and only after that we may do addtions, if any. If there are no parentheses, otherwise the parens dictate the order of evaluation.
What rules govern the order of evaluation for &&, ||, and ! ? ! has the highest priority. Then comes &&, and the || has the lowest priority.
So if you look at 
a || b && ! c
 It's evaluated as 
a || (b && (! c))
What is the truth table for 
!a && !b
 Let's build it at the same time for 
!(a || b)
a b a && b !(a && b) !a !b !a || !b
true true true false false false false
true false false true false true true
false true false true true false true
false false false true true true true
We have just proved one of DeMorgan's law. What is the other one?
It's the dual of this: 
!(a || b)
 is the same as 
!a && !b
There are many other identities that one can prove. Perhaps we can do that later, as needed.
Yes, but let me give some examples, in case you get bored and want to practice. Sure.
This ... is the same as this 
a && (b || c)
 
a && b || a && c
 
a || true
 
true
 
a && true
 
a
 
a || false
 
a
 
a && false
 
false
 
a == true
 
a
 
a == false
 
! a
Let's work some more exercises... ... with if statements.
OK, the text of the problem is always the same: what is the output produced by the following snippet of code... ... when embedded in a complete program. Let's see the snippets.
Snippet 1: 
int x = 3;
if (2 > x) 
  System.out.print(1); 
else 
  System.out.print(2); 
if (x < 2) 
  System.out.println(3); 
System.out.print(4);
 Easy. Draw a diagram.
Snippet 2: 
int x = 3;
if (x > 5) { 
  if (x < 10) 
    System.out.print(1); } 
else 
  System.out.print(2); 
  System.out.print(3);
 Messy. The curly braces change everything.
What if you take them out? The diagram changes significantly.
And you have experienced a dangling else. That's right.
Snippet 3: 
int x = 3; 
if (x > 0) System.out.print(x + 1); 
else if (x > 1) System.out.print(x); 
else if (x > 2) System.out.print(x - 1); 
else if (x > 3) System.out.print(2 * x); 
else System.out.print(x * x);
Easy. Diagram it.
Snippet 4 (and last): 
int x = 3; 
if (x > 0) System.out.print(x + 1); 
else if (x > 1) System.out.print(x); 
else if (x > 2) System.out.print(x - 1); 
else if (x > 3) System.out.print(2 * x); 
else System.out.print(x * x);
Who would ever do that in a program?
Nobody. It's for practice. Messy again. You have to redraw everything.
I agree it's messy, but is it hard? No. Is this the last one?
Yes. Can we do a reasonable example now?
OK, here's problem 19 from problem set 2. 
/* Solution to problem nineteen in the first problem set. Use 
ConsoleReader from lab notes 2 as explained. The trick here (as 
hinted in the text) is to transform a number for a month in a 
position (index) in the string where the month name is starting, 
all names being made of the same length, and then concatenated 
together in one final string. */ 

public class Nineteen { 
    public static void main(String[] args) {

	String monthNames = "January   " +
                            "February  " +
                            "March     " +
                            "April     " +
                            "May       " +
                            "June      " +
                            "July      " +
                            "August    " +
      	                    "September " + // longest
                            "October   " +
                            "November  " +
                            "December  "    ;

	// open a connection with the keyboard 
	ConsoleReader console = new ConsoleReader(System.in); 

	// greet the user, and ask for input 
	System.out.println("Please enter a month number from 1 to 12."); 
	// get month name 
	int month = console.readInt(); 
	// report the name of the month 
	System.out.println(
            monthNames.substring("September ".length() * (month-1), 
                                 "September ".length()*month));

	// formula uses the length of the longest name 
    }
}
Here is it with if statements: 
public class P19 {
    public static void main(String[] args) {

	ConsoleReader console = new ConsoleReader(System.in);

	System.out.println("Please enter a month number from 1 to 12."); 

	int month = console.readInt();

	if      (month == 12) System.out.println("December");
	else if (month == 11) System.out.println("November");
	else if (month == 10) System.out.println("October");
	else if (month ==  9) System.out.println("September");
	else if (month ==  8) System.out.println("August");
	else if (month ==  7) System.out.println("July");
	else if (month ==  6) System.out.println("June");
	else if (month ==  5) System.out.println("May");
	else if (month ==  4) System.out.println("April");
	else if (month ==  3) System.out.println("March");
	else if (month ==  2) System.out.println("February");
	else if (month ==  1) System.out.println("January");
    }   
}
I thought we agreed to use block statements for the bodies of if statements and else alternatives all the time. Yes, but just for once I wanted to keep the code somewhat shorter.
Well, then, just for once, I have two more exercises. OK, I will remember to put braces from now on, always.
Too late. Where's the first snippet?
Here it is: Can't be true! 
if (false && false || true) {
  System.out.print(false); 
} else {
  System.out.print(true); 
}
Snippet 2: I can see the difference. 
if (false && (false || true)) {
  System.out.print(false); 
} else {
  System.out.print(true); 
}
I'm sure you do. That's probably true or false.
Last updated: Feb 4, 2001 by Adrian German for A201