Where you assign a String var in the declaration, not with the new keyword, but creating a new object in the stack just using the quotes;

String s = "this is a string";

1. If both operands are numeric, then the + operation is numeric addition.

2. If either operands is a String, then the + operation is String concatenation.

3. The expression is read left to right.

How would Java concatenation print the following;

int three = 3;

String four = "4";

Syste.out.println(1 + 2 + three + four);

Java would add the ints 1,2, three (3) together first, reading left to right which would be 6. Then it would concatenation the String four.

Lastly it would print out

64.

In Java concatenation, what does the following expression print?

String s = "1";

s += "2";

s += 3;

System.out.println(s);

It would print;

123

+= just means take the previous value, append the new value, and return a new String.

For example

s+=3;

Just means

s = s + 3;

Mutable means _____.

Immutable means _____.

1. Changeable

2. Not Changeable

What does the following print?

String s1 = "1";

String s2 = s1.concat("2");

s2.concat("3");

System.out.println(s2);

The String method _____ returns the number of characters.

The _____ method returns which character is at the specified index.

The _____ method returns firstIndex of the first time the passed in character occurs in the String.

The _____ method returns a String or sequence of characters for the given starting index and (optional) ending index. If the ending index is not included, it is assumed it will move automatically to the end of the String. This method is the trickiest of the String method because it allows you enter an ending index that adds _____ to the last index in the string, or the total character count in the string. Also the first index cannot be _____ than the end index.

1. Length()

2. CharAt(index)

3. indexOf(String)

4. SubString(beginIndex, endIndex)

5. 1

6. More 

False the StringBuilder class is mutable when calling methods on it, the underlying object will reflect the changes made by calling methods on it.

StringBuilder sb = new StringBuilder("this is a ");

sb.append("StringBuilder");

You do not need to return this to an existing reference var or a new reference var.

For a StringBuilder, lets say you had the following scenario;

StringBuilder a = new StringBuilder("a");

StringBuilder b = a.append("b");

What would be the value of variables 'a' and 'b' and why?

The value for both 'a' and 'b' would be "ab".

This is because references 'a' and 'b' are both pointing to the same object, 'a' is the only one with a call to new.

StringBuilder sb1 = new StringBuilder();

StringBuilder sb2 = new StringBuilder("a");

StringBuilder sb3 = new StringBuilder(10);

1. Int for the index to insert, String or other type and value to insert

2. It inserts the value in the second param at the given index

3. A reference to the current StringBuilder

there is also a 4 parameter method, that takes the insert offset, a char[] or CharSequence, and the begin and end index of the char[] or CharSequence

1. Delete and deleteCharAt

2. Delete - int start, int end; deleteCharAt - int index - removes just one char

3. Reference to the current sb.

1. The toString() method

2. Ultimately we are looking to create a string from StringBuilder, and this method returns it.

1. StringBuffer is an older class and was built to be thread-safe.

2. Because it is thread-safe it is slower than StringBuilder

Evaluate the equality of the scenario below and explain it;

StringBuilder sb1 = new StringBuilder("Hello World");

StringBuilder sb2 = sb1;

System.out.println(sb1 == sb2);

Evaluate the following String scenarios;

1. String s1 = "Hello World";

String s2 = "Hello World";

==

2. String s1 = new String("Hello Word");

String s2 = "Hello World";

==

3. String s1 = "Hello World";

String s2 = " Hello World".trim();

==

4. String s1 = "Hello World";

String s2 = " Hello World".trim();

S1.equals(s2)

1. Equals - because they are pointing to the String Literal in the String or intern pool.

2. No equals - One is in JVM String pool, one is a new object in the heap.

3. No equals - while both are String literals, one is evaluated on compile time and one is evaluated at runtime, and are pointing to difference objects, because the runtime object was created internally with new.

4. Equals - the String equals() method looks at compares the internal characters in the String to see if they are equal.

1. Not one that is implemented, so no.

2. It will only return the same referential equality result as ==.

int[] numbers = new int[3];

int numbers[] = new int[]{1,2,3};

int[] numbers = {1,2,3};

Yes it is legal to put a space after the type and reference var name.

int [] numbers;

int numbers [];

What is the difference between the following code?

1. int[] ids, Nums;

2. int ids[], Nums;

1. Is declaring to two arrys of type int.

2. Is declaring 1 array of ints and one int.

Identify the different components of the following;

[Ljava.lang.String;@ 160bc7c0

1. [L means that the object is an array

2. Java.lang.String; means it is an array of Strings

3. @ 160bc7c0 is the hascode for the array

What are the values of the following?

1. String[] stars;

2. String[] stars2 = new String[2];

1. Stars refers to null.

2. Stars2 refers to a String array that has to two references to null.

What does the following print and why?

String wary [] = new String[8];

System.out.println(wary.length);

When looping through an array, should you ever have <= in the comparison statement? Why or why not?

i.e.

int[] numbers = new int[10];

for(int i = 0; i<= numbers.length;++i){

}

If you sort the following array, what will be the result?

String[] Nums = {"10", "90", "100"};

What has to happen with array before you execute a search on it? How do you execute a search for a value in an array? What happens when the value is not found?

i.e.

int[] Nums = {1,2,3,6,8};

what value is returned when we search for 9?

1. You need to sort it (java.util.Arrays.sort(nums)) or the results will be unpredictable.

2. Java.util.Arrays.binarySearch(Nums, value)

3. You find the index where the value should be, negate it and add -1 to that value

4. So the index should be 5, we negate it to -5 and add -1 which is -6.

True or false - When using a multi-dimensional array, you can initialize an arrays first dimension, and then define the size of each array component after in a separate statement.

int[][] numarys = new int[4][];

numarys[0] = new int[4];

numarys[1] = new int[2]; 

ArrayList<String> strings = new ArrayList<String>();

ArrayList<String> strings = new ArrayList<>();

ArrayList list1= new ArrayList();

ArrayList list2 = new ArrayList(3);

ArrayList list3 = new ArrayList(list2);

boolean add(Object) (primitives can be autoboxed) always returns true

void add(int index, E element) E stands for whatever class. You want to put in the ArrayList

E set(index i, E type)

It will generate an error

1. IsEmpty()

2. Size()

Boolean.parseBoolean("TrUe");

Short.parseShort("0");

Byte.parseByte("1");

Integer.parseInteger("2");

Long.parseLong("3");

Float.parseFloat("4");

Double.parseDouble("4.0");

Character does not have a parse method 

Booealn.valueOf("True");

...Double.valueOf("4.89");

All wrapper types have valueOf methods, except Character.

List<String> l = new ArrayList<>();

1. String[] ary1 = l.toArray();

2. String [] ary2 = l.toArray(new String[0]);

Explain the concept of an array backed list?

What happens when you try to remove an element from the List?

1. When you want to convert from an array to a list, you exectute;

String[] oldAry = new String[4];

List newList = Arrays.asList(oldAry);

When you do this you have created an fixed-size, array backed list - when you make a change to one, the other will be affected as well.

2. Because an array cannot change its size on the fly, meaning it cannot add or remove elements, the list cannot change either.

The asLIst() method takes a Varargs param Arrays.asList("one", "two", "three");

This is handy for testing and creating a list on one line.

LocalDate ld = LocalDate.now();

LocalTime lt = LocalTime.now();

LocalDateTime ldt = LocalDateTime.now();

2015-01-02

T12:35:34.401

2015-01-02T12:35:34.401

LocalDate ld1 = LocalDate.of(2015, Month.JANUARY, 7);

LocalTime lt1 = LocalTime.of(11, 36, 59, 200);

LocalDateTime ldt1 = LocalDateTime.of(ld1, lt1);

OR

LocalDateTime ldt2 = LocalDateTime.of(2015, Month.JANUARY, 7, 11, 36, 59, 201);

.plusDays(int),

.plusYears(int),

.plusMonths(int),

.plusMinutes(int),

.plusHours(int),

.plusSeconds(int),

and the minus versions of these methods

REMEMBER YOU MUST ASSIGN THE VALUE BACK TO AN OBJECT REFERENCE

True or false - you can use object methods from LocalDate and LocalTime on their respective objects. In other words, it is legal to call a method like plusDays() on a LocalTime object, and minusSeconds() on a LocalDate object, and it will compile.

i.e.

LocalTime lt1 = LocalTime.now();

lt1 = lt1.plusDays(1);

False, you can only use the related methods to the fields that are associated with that object.

i.e.

LocalTime lt1 = LocalTime.now();

lt1 = lt1.plusMinutes(1);

Period months = Period.ofMonths(4);

Period years = Period.ofYears(3);

Period days = Period.ofDays(5);

Period weeks = Period.ofWeeks(6);

Period yearAndAWeek = Period.of(1, 0, 7);

True or false - method chaining is legal for Period class;

Period p = Period.ofMonths(2).ofDays(6);

True or false - you can mix date and time Period info among LocalDate and LocalTime classes.

Period months = Period.ofMonths(3);

LocalTime time = LocalTime.now();

time.plus(months);

Use the built-in format method by passing in the DateTimeFormatter type.

LocalTime time = LocalTime.now();

String tStr = time.format(DateTimeFormatter.ISO_LOCAL_TIME);

Or ISO_LOCAL_DATE, Or ISO_LOCAL_DATE_TIME

DateTimeFormatter.ISO_LOCAL_DATE = "2012-01-01"

DateTimeFormatter.ISO_LOCAL_TIME = "11:35:46.890"

DateTimeFormatter.ISO_LOCAL_DATE_TIME = "2012-01-01T11:35:46.890"

True - for example;

LocalDate ld = LocalDate.now(2013, Month.JANUARY, 29);

String dateStr = ld.format(DateTimeFormatter.ISO_LOCAL_DATE);

AND

DateTimeFormatter dtf = DateTimeFormatter.ofLocalizedDate(FormatStyle.SHORT);

String dateStr2 = dtf.format(ld);

1. 01/01/2012

2. 11:12 AM

3. 01/01/2012 11:12 AM

1. Jan 20, 2013

2. 11:12:45 AM

3. Jan 20, 2013 11:12:45 AM

Provide an example of the DateTimeFormatter.ofPattern() method.

LocalDateTime dateTimeNow = LocalDateTime.now();

DateTimeFormatter dtf = DateTimeFormatter.ofPattern("MMMM, yyyy, dd");

String dateTimeStr = dtf.format(dateTimeNow);

Use the parse() method on the objects.

LocalDate date = LocalDate.parse("01-01-2012");

OR

DateTimeFormatter dtf = DateTimeFormatter.ofPattern("MM dd yyyy");

date = LocalDate.parse("01-01-2013", dtf);

True or false = when instantiating an array like;

String[] strAry = new String[]{"1", "2"};

it is also legal to instantiate by moving the brackets on the right before String and after new.

True or false - when using the diamond operator as a generic declaration element, you have the option to only include it on the left side of the declaration.

i.e.

List<Integer> l = new ArrayList();

True

List l = new ArrayList();

List l2 = new ArrayList();

l.add("hello");

l2.add("goodbye");

List l3 = l2.subList(0, 1); // this is the proper way to get a sublist

l.add(l3);

List l4 = l2.subList(0,0); // this returns an empty list.

True or false - you can initialize an array by using the array size and an anonymous array on the right side of the declaration.

String[] sA = new String[1] {"aaa"};

True or false - String builder does not have a 4 parameter insert method that inserts a portion of a charsequence into the designated offset char;

i.e.

public StringBuilder insert(int dstOffset, CharSeqence s, int start, int end)