1.15 String Manipulation

1. Creating String objects

A String object represents a sequence of characters. You can create one with a string literal or by calling the String class constructor 1.15.A.1.

The String class is part of the java.lang package, and classes in java.lang are available by default 1.15.A.2, so no import is needed.

String team = "Falcons";             // string literal
String city = new String("Porto");   // String constructor
System.out.println(team + " of " + city);

Output:

Falcons of Porto

2. Strings are immutable

A String object is immutable: after it is made, nothing about it can change, and calling a method on it never changes its characters 1.15.A.3. Methods such as substring return a new String; no String method changes the original.

String w = "data";
w.substring(1);           // new String made, but not stored
System.out.println(w);
w = w.substring(1);       // now w refers to the new String
System.out.println(w);

Output:

data
ata

Watch out: a String method result that is not stored is lost.

3. Concatenation with + and +=

Two String objects can be concatenated (joined) with the + or += operator, which results in a new String object. A primitive value can be concatenated with a String; the primitive is implicitly converted to a String 1.15.A.4. Java works left to right, so where the text first appears matters.

Code	Value or output	Why
"Score: " + 4 + 5	"Score: 45"	left to right: text first, so 4 and 5 are joined as text
4 + 5 + " pts"	"9 pts"	left to right: 4 + 5 is added first
w.substring(1); then print w (w is "data")	data	the new String was not stored
w = w.substring(1); then print w	ata	now the variable refers to the new String

System.out.println("Score: " + 4 + 5);
System.out.println(4 + 5 + " pts");
String msg = "Level";
msg += " " + 3;
msg += "!";
System.out.println(msg);
boolean won = true;
System.out.println("Won: " + won);

Output:

Score: 45
9 pts
Level 3!
Won: true

4. Concatenating objects: toString

A String can be concatenated with any object. This implicitly calls the object’s toString method, which returns a string representing the object. Every object has a toString method, because every class is a subclass of Object (topic 1.12) and inherits it 1.15.A.5.

Subclasses of Object often override toString with their own class-specific version. A subclass overrides a method (method overriding) when it writes a public method with the same method signature as a public method of its superclass, so that the method behaves in a way that fits the subclass 1.15.A.5.

Example: p refers to a Plant whose toString method returns "Fern, 30 cm".

Plant p = new Plant("Fern", 30);
System.out.println("Plant: " + p);
p.grow(2);
String label = "Now: " + p;
System.out.println(label);

Output:

Plant: Fern, 30 cm
Now: Fern, 32 cm

Outside the course: writing (overriding) your own toString method. You only need to know that concatenation calls it.

5. Indices and StringIndexOutOfBoundsException

A String has index values from 0 to one less than its length. Attempting to access an index outside this range results in a StringIndexOutOfBoundsException 1.15.B.1. (The to argument of substring may equal length(), because it stops before to. The from argument may also equal length(): "moon".substring(4) is "".)

For String word = "INTERFACE";:

Index	0	1	2	3	4	5	6	7	8
Character	I	N	T	E	R	F	A	C	E

• length() is 9, so the valid indices are 0 to 8.
• word.substring(0, 5) is "INTER": it stops BEFORE index 5.
• word.substring(5) is "FACE": from index 5 to the end.
• word.substring(2, 3) is "T".

String word = "INTERFACE";
System.out.println(word.substring(0, 5));
System.out.println(word.substring(5));
System.out.println(word.substring(2, 3));
System.out.println("moon".substring(4).length());
System.out.println(word.substring(7, 10));

Output:

INTER
FACE
T
0
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: Range [7, 10) out of bounds for length 9

Index 10 is past the end (the largest allowed to is 9), so the last call throws the exception. The message after the colon can differ between Java versions.

6. The six String methods on the Java Quick Reference

These String methods, what they do and when to use them are part of the Java Quick Reference 1.15.B.2:

Method	Returns	Example with word = "INTERFACE"
int length()	the number of characters	word.length() is 9
String substring(int from, int to)	the characters from index from to to - 1	word.substring(0, 5) is "INTER"
String substring(int from)	substring(from, length())	word.substring(5) is "FACE"
int indexOf(String str)	the index of the first occurrence of str; -1 if not found	word.indexOf("FACE") is 5; word.indexOf("X") is -1
boolean equals(Object other)	true if the same sequence of characters, false otherwise	"Cat".equals("cat") is false
int compareTo(String other)	a value < 0 if this comes first alphabetically, 0 if equal, a value > 0 if this comes later	"cat".compareTo("dog") < 0 is true

Outside the course: using equals to compare a String with an object that is not a String.

String word = "INTERFACE";
System.out.println(word.length());
System.out.println(word.indexOf("FACE"));
System.out.println(word.indexOf("E"));
System.out.println(word.indexOf("X"));
System.out.println("Cat".equals("cat"));
System.out.println("cat".compareTo("dog") < 0);
System.out.println("pear".compareTo("pear"));
System.out.println("plum".compareTo("pear") > 0);

Output:

9
5
3
-1
false
true
0
true

indexOf("E") gives 3, the first E, even though there is another E at index 8.

Watch out: compare String contents with equals, not ==. The == operator checks whether two references point to the same object:

String a = "go";
String b = new String("go");
System.out.println(a.equals(b));
System.out.println(a == b);

Output:

true
false

a and b hold the same characters, but they refer to two different objects.

7. One character, empty strings and null

A string identical to the single character at position index can be created with substring(index, index + 1) 1.15.B.3. The last character is s.substring(s.length() - 1).

String s = "ROBOT";
System.out.println(s.substring(2, 3));
System.out.println(s.substring(s.length() - 1));
int pos = s.indexOf("B");
System.out.println(s.substring(pos, pos + 1));

Output:

B
T
B

Empty string or no string? "" is the empty string: a real String object with length 0, so "".length() is 0, and s.substring(i, i) returns "" for any i from 0 to s.length(). null means no object at all: if s is null, s.length() throws a NullPointerException 1.14.A, 1.15.B.

String empty = "";
System.out.println(empty.length());
System.out.println("ROBOT".substring(3, 3).length());
String none = null;
System.out.println(none.length());

Output:

0
0
Exception in thread "main" java.lang.NullPointerException: Cannot invoke "String.length()" because "none" is null

Background (not tested): characters and capitals. Java also has a char type ('A') and methods such as charAt and toUpperCase. char is outside the course and those methods are not on the Java Quick Reference, so use substring in exam answers. compareTo compares character codes, so all capital letters come before lowercase ones: "Zoo".compareTo("apple") is negative. On the exam, strings are “ordered based upon the alphabet”, so most questions use the same case throughout.

System.out.println("Zoo".compareTo("apple") < 0);

Output:

true