Chapter 1: Introduction to Computers, Programs, and Java
Q: When you compile your program, you receive an error as follows:
%javac Welcome.java
javac not found
What is wrong?
A: Two possible reasons: 1. Java is not installed on your system; or 2. Java is installed, but the
environment variable PATH is not set. To set the path correctly, see Section 1.3 in
Supplement I.B.
Q: When you compile your program, you receive an error as follows:
%javac Welcome.java
javac: file not found: Welcome.java
Usage: javac
use -help for a list of possible options
What is wrong?
A: File Welcome.java does not exist in the directory. Check if Welcome.java is in the current directory.
Q: When you run your program, you receive an error as follows:
%java Welcome
Exception in thread "main" java.lang.NoClassDefFoundError: Welcome
What is wrong?
A: File Welcome.class does not exist. You have to first compile Welcome.java to produce Welcome.class in the current directory.
Q: When you run your program, you receive an error as follows:
%java Welcome
Exception in thread "main" java.lang.NoSuchMethodError: main
What is wrong?
A: Please check if you have a main method in Welcome.java or you have spelled the method public static void main(String[] args) correctly.
Q: When you compile your program, you receive an error as follows:
%javac Welcome.java
Welcome.java:2: cannot find symbol
symbol : class string
location: class Welcome
public static void main(string[] args) {
^
1 error
What is wrong?
A: Java is case-sensitive. string should be spelled as String.
Chapter 2:
Elementary Programming
Q: What is wrong in the following code?
public class Test {
public static void main(String[] args) {
i = 1;
j = i * i * i;
}
}
A: You will get a compile error that indicates symbol i cannot be resolved. Every variable must be declared before being used.
Q: What is wrong in the following code?
1 public class Test {
2 public static void main(String[] args) {
3 int i;
4 j = i * i * i;
5 }
6 }
A: You will get a compile error that indicates variable i is not initialized in line 3.
Q: What is wrong in the following code?
1 public class Test {
2 public static void main(String[] args) {
3 int i;
4 int j = 5;
5 System.out.println(j * 4.5);
6 }
7 }
A: The program compiles and runs fine. However, i is never used. This might be a potential programming error. Most IDE now can alert you of this situation.
Q: Why does this program display 0?
1 public class Test {
2 public static void main(String[] args) {
3 int i = 9;
4 int j = 5 / 6 * (i * 4);
5 System.out.println(j);
6 }
7 }
A: This type of mistakes is very common for my students. We have covered this repeatedly in the text.
Here just to remind you again 5 / 6 is 0, because both 5 and 6 are integers.
Q: What is the output of this program?
public class Test {
public static void main(String[] args) {
System.out.println(2147483647 + 1);
}
}
A: Run this program. You will see the output is -2147483648. Why? 2147483647 is the largest int value. Adding 1 to this value causes overflow.
It is not a runtime error in Java.
Q: What is the output of the following code?
1 public class Test {
2 public static void main(String[] args) {
3 System.out.println("1 + 2 is " + 1 + 2);
4 }
5 }
A: The program displays 1 + 2 is 12. If you want the output to be 1 + 2 is 3, put
1 + 2 inside the parentheses (1 + 2) to force 1 + 2 to
be evaluated first.
Q: When you write the following code in Eclipse or NetBenas, you may get a warning "resource leak" on input. Why?
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter an integer: ");
int v = input.nextInt();
System.out.println("You entered " + v);
}
}
A: Ignore this warning. It will not cause any problem. Java recommends
that you close the input explicitly by invoking the close() method
on input to prevent potential memory leak. We will cover this in
Chapter 12. In this case, there is no memory leak, because the input
object is discarded when your program is finished.
Q: Identify potential problems in the following code.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter an integer: ");
int v = input.nextInt();
System.out.println("You entered " + v);
Scanner input2 = new Scanner(System.in);
System.out.print("Enter a double: ");
double v2 = input2.nextInt();
System.out.println("You entered " + v2);
}
}
A: The code creates new Scanner(System.in) twice. This is unnecessary and
it is a bad code. The code will not work with input redirection.
Suppose you want to use input redirection to read data from file
Temp.txt using the following command:
java Test < Temp.txt
Suppose Temp.txt contains
3
3.5
You will get an error. To fix it, rewrite the code as follows:
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter an integer: ");
int v = input.nextInt();
System.out.println("You entered " + v);
System.out.print("Enter a double: ");
double v2 = input.nextInt();
System.out.println("You entered " + v2);
}
}
Chapter 3: Selections
Q: What is wrong in the following code?
1 public class Test {
2 public static void main(String[] args) {
3 int i, j, k;
4 i = j = k = 3;
5 if (i < j < k)
6 System.out.println("i, j, and k are in increasing order");
7 }
8 }
A: i < j evaluates to a boolean value, which cannot be compared with k. The correct expression should be (i < j && j < k).
Q: What is the output of the following code?
public class Test {
public static void main(String[] args) {
System.out.println(1.0 - 0.1 - 0.1 - 0.1 - 0.1 - 0.1 == 0.0);
}
}
A: Run the program. You will be surprised to see the output is false. Why? because floating-point numbers are not stored in
complete accuracy. You are dealing with approximation for floating-point numbers.
Don't compare two floating-point numbers using equality == operators.
Q: What is the output of the following code?
public class Test {
public static void main(String[] args) {
boolean done = false;
if (done = true) {
System.out.println("done");
}
}
}
A: Run the program. You will see the output is done. Why? you assign true to done. So, the condition is true. To check
whether done is true, you should use done == true, or simply
if (done) {
System.out.println("done");
}
Q: What is the output of the following code?
public class Test {
public static void main(String[] args) {
boolean done = false;
if (done);
{
System.out.println("done");
}
}
}
A: Run the program. You will see the output is done. Why? because the if statement ends with
if (done);
The semicolon ends the if statement. The lines
{
System.out.println("done");
}
are not part of the if statement.
Chapter
4: Mathematical Functions, Characters, and Strings
Q: What is the output of this program?
public class Test {
public static void main(String[] args) {
char ch = 'a';
System.out.println('a' + 1);
System.out.println(++ch);
}
}
A: for 'a' + 1, 'a' is converted into int. so 'a' + 1 is 97 + 1 = 98. For
++ch, the result is 'b'.
Q: What is the output of the following code?
public class Test {
public static void main(String[] args) {
char ch = 'a';
switch (ch) {
case: 'a': System.out.print(ch);
case: 'b': System.out.print(ch);
case: 'c': System.out.print(ch);
}
}
}
A: Run the program. You will see the output is aaa. Why? There is no break statement for the case statement.
Once a case is matched, the statements starting from the matched case are executed until a break
statement or the end of the switch statement is reached.
This phenomenon is referred to as the fall-through behavior.
Chapter 5: Loops
Q: What is wrong in the following code?
public class Test {
public static void main(String[] args) {
int i = 0;
while (i < 10) {
System.out.println("Welcome to Java!");
}
}
}
A: i is always 0. This is an infinite loop.
This is a common error for while and do-while loops. Make sure the control variable changes
to cause the loop to stop eventually.
Q: What is wrong in the following code?
public class Test {
public static void main(String[] args) {
int i = 0;
while (i < 10)
System.out.println("Welcome to Java!");
i++;
}
}
A: i++ is not in the body of the while loop. So, this is an infinite loop. If the loop body contains more than
one statement, use the braces to group them.
Chapter 6: Methods
Q: What is the output of the following code?
public class Test {
public static void main(String[] args) {
int x = 1;
p(x);
System.out.println("x is " + x);
}
public static void p(int x) {
x = x + 1;
}
}
A: x is 1. Invoking p(x), the value of x is passed to the parameter in the method. Variable x defined in
the main method and x defined in method p are two independent variables.
Q: Does Math.sin(90) return the sine value for angle 90 degree?
A: No. To return the sine value for angle 90 degree, use Math.sin(Math.PI / 2).
Q: Identify the potential issues in the following code.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
for (int i = 0; i < 2; i++) {
System.out.println(average());
}
}
public static double average() {
Scanner input = new Scanner(System.in);
System.out.print("Enter five numbers: ");
double sum = 0;
for (int i = 0; i < 5; i++)
sum += input.nextDouble();
return sum / 5;
}
}
A: At first glance, the code is fine and it runs fine. However, if you run the code by entering 10 numbers in one line,
the program will not receive these input for the second iteration. The program will not work if you use input redirection
as follows:
java Test < Temp.txt
Enter five numbers: 3.0
Exception: NoSuchElementException
Suppose Temp.txt contains numbers \
1 2 3 4 5
6 7 8 9 10
To fix the problem, declare input as static variable outside the method as follows:
import java.util.Scanner;
public class Test {
static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
for (int i = 0; i < 2; i++) {
System.out.println(average());
}
}
public static double average() {
System.out.print("Enter five numbers: ");
double sum = 0;
for (int i = 0; i < 5; i++)
sum += input.nextDouble();
return sum / 5;
}
}
In fact, this is the preferred way to declare and create a Scanner object for input. If you read data from
a method other than the main method, declare and create Scanner outside the method.
Q: My program output is correct, but the grader marked my method wrong, why?
A: A common error for the exercises in this chapter is that students don't implement the methods to meet the
requirements even though the output from the main program is correct.
For an example of this type of error see www.cs.armstrong.edu/liang/CommonMethodErrorJava.pdf.
Chapter 7:
Single-Dimensional Arrays
Q: What is wrong in the following code?
public class Test {
public static void main(String[] args) {
int[] x = new int[10];
for (int i = 0; i <= x.length; i++)
System.out.print(x[i]);
}
}
A: This is a common error, known as off-by-one. Here the array index is from 0 to 9, but the last i in the loop
will be x.length (10). At 10, the index is out of range.
Q: What is wrong the following code?
public class Test {
public static void main(String[] args) {
int[] x = new int[10];
p(x);
System.out.print(x[0]);
}
public static void p(int[] list) {
list[0] = 10;
}
}
A: The output is 10. When invoking p(x), the reference value of array x is passed to list. The first element in the array
is changed in the method.
Chapter 9: Objects and
Classes
Q: Do local variables have default values? Do instance variables have default values?
Do static variables have default values?
A: Local variables don't have default values. Instance and static
variables are data fields and have default values.
Q: What is the printout of the following code?
public class Test {
public void Test() {
System.out.println("Test");
}
public static void main(String[] args) {
Test t = new Test();
}
}
A: None. Note that "public void Test()" is not a constructor. It is a method. The class Test has a default constructor.
Q: What is wrong the following code?
public class Test {
public static void main(String[] args) {
String[] x = new String[10];
System.out.print(x[0].toString());
}
}
A: When you create an array of objects. Each element of the array
has a null value. So, invoking x[0].toString() causes a
NullPointerException.
Chapter 11: Inheritance and
Polymorphism
Q: What is wrong the following code?
public class Apple extends Fruit {
}
class Fruit {
public Fruit(String name) {
System.out.println("Fruit's constructor is invoked");
}
}
A: Since no
constructor is explicitly defined in Apple, Apple’s
default no-arg constructor is defined implicitly. Since Apple
is a subclass of Fruit, Apple’s default constructor
automatically invokes Fruit’s no-arg constructor. However,
Fruit does not have a no-arg constructor, because Fruit
has an explicit constructor defined. Therefore, the program cannot
be compiled.
Q: What is the difference between == and equals()?
A:
The ==
comparison operator is used for comparing two primitive data type
values or for determining whether two objects have the same
references. The equals method is intended to test whether two
objects have the same contents, provided that the method is
overridden in the defining class of the objects. The ==
operator is stronger than the equals method, in that the == operator checks whether the two reference variables refer to
the same object.
Chapter 12:
Exception Handling and Text I/O
Q: What is wrong in the following code?
import java.util.Scanner;
public class Test2 {
public static void main(String[] args) {
for (int i = 0; i < 1; i++) {
System.out.println(average());
}
}
public static double average() {
double sum = 0;
try (Scanner input = new Scanner(System.in);) {
System.out.print("Enter five numbers: ");
for (int i = 0; i < 5; i++)
sum += input.nextDouble();
}
return sum / 5;
}
}
A: The
try-with-resource syntax automatically closes the input. It also
closes System.in by invoking System.in.close(). So you will get an
error when the average method is invoked again. To fix it, write the
code as shown in Question 3 in Chapter 6 on this page.
Chapter 18: Recursion
Q: What is wrong the following code?
public static long factorial(int n) {
return n * factorial(n - 1);
}
A: There is
no stopping condition. The method runs infinitely.
Q: What is wrong the following code?
public class Test {
public static void main(String[] args) {
printLine(2, "Welcome to Java");
}
public static void printLine(int n, String message) {
while (n > 1) {
System.out.println(message);
printLine(n - 1, message);
}
}
}
A: while
should be replaced by if. With "while", the method runs infinitely.
To understand why this happens, draw a stack. First the main method
is placed into the stack. Second, printLine(2, "Welcome to Java") is
placed into the stack. Now n is 2. Third, printLine(1, "Welcome to
Java") is placed into the stack. Now n is 1, so the method finished
and popped out of the stack. But now you are still in the second
method call. In the second method call, n is still 2. The
loop-continuation-condition is still true. So, the loop continues.
The method will run forever.
