Except where otherwise noted, the contents of this document are Copyright 2019 Stuart Reges and Marty Stepp.
lab document created by Marty Stepp, Stuart Reges and Whitaker Brand
Goals for this problem set:
boolean
expressions and variables to represent logical true/false expressionsboolean
type
The boolean
type represents logical values of true
or false
. Combine boolean
expressions with logical operators &&
(and), 
(or), and !
(not).
Example:
boolean test1 = 7 < 10; // true boolean test2 = (1 == 2); // false if ((test1  test2) && 2 + 2 != 5) { System.out.print("hello"); // output: hello }
String
methods with boolean
resultsMethod name  Description 

string.equals(string)

whether the two strings are identical 
string.equalsIgnoreCase(string)

whether the two strings are identical, ignoring capitalization 
string.startsWith(string)

whether this string begins with the characters of the given string 
string.endsWith(string)

whether this string ends with the characters of the given string 
string.contains(string)

whether the characters of the given string occur within this string 
String name = "Professor Smith"; if (name.startsWith("Prof")) { System.out.println("When are your office hours?"); }
Write the result of each expression as either true
or false
, given the following variables.
int x = 12; int y = 7; int z = 28; String s = "mid term";
x < 14 
true 

!(x % 2 < 1) 
false 

x < y  x < z 
true 

z / x < x / y * x 
true 

s.length() == y 
false 

s.toUpperCase().equals("MID TERM") 
true 

!s.equals("mid term")  x * y != z 
true 

s.substring(z / x).length() > y 
false 
Write a method named allDigitsOdd
that returns whether every
digit of a positive integer is odd. Your method should
return true
if the number consists entirely of odd digits
and false
if any of its digits are even. 0, 2, 4, 6, and 8
are even digits, and 1, 3, 5, 7, 9 are odd digits.
For example, allDigitsOdd(135319)
returns true
but allDigitsOdd(9145293)
returns false
.
Hint: You can pull apart a number into its digits using /
10
and % 10
.
Write a method hasMidpoint
that accepts three
integers as parameters, and returns true
if one of the numbers
is the midpoint of the other two and returns false
otherwise.
For example, the call hasMidpoint(3, 7, 5)
would
return true
because one of the parameters (5) is the midpoint
of the other two (3 and 7).
Try to solve this problem in PracticeIt: click on the checkmark above!
Write a method before
that takes as parameters two month/day
combinations and that returns whether or not the first date comes before
the second date (true
if the first month/day comes before the
second month/day, false
if it does not). The method will take
four integers as parameters that represent the two month/day combinations.
The first integer in each pair represents the month and will be a value between 1 and 12 (1 for January, 2 for February, etc, up to 12 for December). The second integer in each pair represents the day of the month (a value between 1 and 31). One date is considered to come before another if it comes earlier in the year.
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Write a method sameDashes
that takes two strings as parameters
and that returns whether or not they have dashes in the same places
(returning true
if they do and returning false
otherwise). For example, below are four pairs of strings of equal length
that have the same pattern of dashes. Notice that the last pair has no
dashes at all.
string 1: "hithereyou." "15389" "criminalplan" "abc" string 2: "12(134)7539" "xyzzy" "(206)5551384" "9.8"To be considered a match, the strings must have exactly the same number of dashes in exactly the same positions. The Strings might be of different length.
Solve this problem in PracticeIt by clicking on the checkmark above.
This attempted solution to SelfCheck 5.15 (isVowel
) has
several problems:
// Returns whether the given string represents a vowel: // a, e, i, o, or u, case insensitively. public static boolean isVowel(String s) { if (s == "a") { return true; } else if (s == "e") { return true; } else if (s == "i") { return true; } else if (s == "o") { return true; } else if (s == "u") { return true; } else { return false; } }
Open PracticeIt from the link above, copy/paste this code into it, then see the next slide.
Fix the following aspects of the code:
public static boolean isVowel(String s) { s = s.toLowerCase(); if (s.equals("a")  s.equals("e")  s.equals("i")  s.equals("o")  s.equals("u")) { return true; } else { return false; } }
The above can be improved. "Boolean Zen" version:
public static boolean isVowel(String s) { s = s.toLowerCase(); return s.equals("a")  s.equals("e")  s.equals("i")  s.equals("o")  s.equals("u"); }
Identify whether each assertion is always/never/sometimes true
at each point.
x > y 
z == 0 
x == y 


A  SOMETIMES  ALWAYS  SOMETIMES 
B  SOMETIMES  SOMETIMES  NEVER 
C  ALWAYS  NEVER  NEVER 
D  NEVER  NEVER  NEVER 
E  NEVER  SOMETIMES  ALWAYS 
public static void mystery(int x, int y) { int z = 0; // Point A while (x != y) { // Point B z++; if (x > y) { // Point C x = x / 10; } else { // Point D y = y / 10; } } // Point E System.out.println(x + " " + y + " " + z); }
You can also solve this problem in PracticeIt by clicking on the checkmark above.