CSE1710. Big Picture. Today is last day covering Ch 6. Tuesday, Dec 02 is designated as a study day. No classes. Thursday, Dec 04 is last lecture.

Size: px

Start display at page:

Download "CSE1710. Big Picture. Today is last day covering Ch 6. Tuesday, Dec 02 is designated as a study day. No classes. Thursday, Dec 04 is last lecture."

Bonnie Ball
6 years ago
Views:

1 CSE1710 Click to edit Master Week text 12, styles Lecture 23 Second level Third level Fourth level Fifth level Fall 2014! Thursday, Nov 27, Big Picture Today is last day covering Ch 6 Tuesday, Dec 02 is designated as a study day. No classes. Thursday, Dec 04 is last lecture. 2 1

2 Revisiting the String class We discussed many important methods from the String class int : length() char: charat(int) String : substring(int,int),string : substring(int) int : indexof(string), int : indexof(string,int) String : tostring(); boolean : equals() int : compareto(string) String : touppercase(), String : tolowercase() 3 boolean : matches(string) String[] : split(string) String : replaceall(string) RELATED CLASSES: StringBuffer, wrapper classes Comparing strings: equals vs matches! suppose we have two strings, str1 and str2!! str1.equals(str2) returns true iff! str1 has the same state as str2! str1.matches(str2) returns true iff! str1 matches the pattern as stipulated by str2!! FOR NOW, WE WILL DO DEAD SIMPLE PATTERNS 4 4 2

3 hello.matches( hello )!! in the context of being a parameter to matches, str2 is interpreted as a regular expression (aka REGEX)!! the REGEX specifies 5 criteria: hello.matches( hello )! REGEX criteria! Criterion satisfied?! that the character h is in index position 0 that the character e is in index position 1 that the character l is in index position 2 that the character l is in index position 3 that the character o is in index position 4 (implied) no further characters in the sequence 5 5 Regular expressions: Simple classes!! a regular expression can also use special characters and syntax to specify more patterns more generally! [abc] defines a simple class of characters REGEX criteria! hello.matches( [Hh]ello )! the character H or h is in index position 0 the character e is in index position 1 the character l is in index position 2 the character l is in index position 3 the character o is in index position 4 no further characters in the sequence str1 satisfies?! L17App

4 Regular expressions: Simple classes using a range!! [a-d] defines a simple class using a range REGEX criteria! hello.matches( [a-d]ello )! the character a or b or c or d is in index position 0 the character e is in index position 1 the character l is in index position 2 the character l is in index position 3 the character o is in index position 4 no further characters in the sequence str1 satisfies?! L17App3 7 7 Regular Expressions!! [a-d[f-h]] matches! any of a,b,c,d,f,g,h! the union of a-d and f-h!! [^a-d] matches! any character that is NOT a, b, c, d,! \d matches any digit! same as: [0-9]!! \s matches any whitespace character:! same as: [ \t\n\x0b\f\r]!! vertical tab is \xob, aka \u000b!! \w matches any word character:! same as: [a-za-z_0-9]! 8 8 4

5 Regular Expressions!! a* matches! zero or more a s! a+ matches! 1 or more a s! a? matches! 0 or 1 a s! a{n,m} matches! at least n a s but not more than m a s 9 9 Regular Expressions 10 Suppose we wish to validate a string representing a time, which must be one of 3, 6, or 9 am or pm! acceptable: 9 am, 3 pm! not acceptable: 10 am, 3 um, 9am, 9:00 am! construct a regex to match this! [369] [ap]m! suppose we want to allow the space to be optional! acceptable: 9am, 12 am, 12pm! not acceptable: 10am, 9:00am! construct a regex to match this! [369]?[ap]m or [369][ ]?[ap]m! 10 5

6 The method String[] : split(string) asdjslkad Example String str = Here is a string! ; String[] tokens; // this declares a variable that has the type array of String elements tokens = str.split( ); // here we invoke the split method, which returns an array of String elements, we assign the RHS to the variable we already declared int numtokens = tokens.length; // here we determine the number of elements in the array 12 6

7 Example // here we iterate over the elements of the array // using a for loop for (int index = 0; index < numtokens; index++) { } System.out.println(tokens[index]); 13 Example // here we iterate over the elements of the array // using collection-based iteration for (String s : tokens) { } System.out.println(s); 14 7

8 StringBuffer, a really cool class Examples: StringBuffer buf1 = new StringBuffer( Hi ); buf1.append( ); buf1.append( There! ); buf1.append( \n ); StringBuffer buf2 = new StringBuffer(!); buf2.insert(0, Hi); buf2.insert(2, There!!); buf2.insert(2, ); buf2.append(\n); buf2.delete(8,10); 15 StringBuffer, a really cool class Examples: StringBuffer buf3 = new StringBuffer( Notification ); buf3.reverse(); System.out.println(buf3.toString()); 16 8

9 state? attributes? object is mutable? (state can be changed) has mutator methods? objects can be operands with + operator? 17 String sequence of characters NO! NO! YES! can masquerade as primitive operand can also invoke methods on object reference instantiation? standard way String s = new String( Hi ); also can use way that masquerades as primitive String s = Hi ; StringBuffer sequence of characters YES! YES! append(string) delete(int, int) insert(int, String) reverse() NO! can only invoke methods on object reference only the standard way StringBuffer s; s = new StringBuffer( Hi ); String instantiation: differences? We have two ways to instantiate string objects String s1 = new String( Hi ); String s2 = Hi ; Are these two ways actually identically the same? 18 9

10 String instantiation: differences? We have two ways to instantiate string objects String s1 = new String( Hi ); String s2 = Hi ; Are these two ways actually identically the same? actually. not exactly the same (next example will illustrate) 19 String instantiation via constructor String s1 = new String( Hi ); String s2 = new String( Hi ); String s3 = new String( Hi ); With the regular, old constructor approach, these statements will result in the creation of 3 different String objects at run time. The objects do happen to have the same state, but they are indeed different objects

11 String instantiation via shortcut String s4 = Hi ; String s5 = Hi ; String s6 = Hi ; With the shortcut approach, these statements will result in the creation of only one String object at run time. The shortcut approach will reuse a String object if one already exists with the required state. 21 There is a pool at run time to keep track of String objects created via this shortcut method. and while we re at it Let s take another look at Strings masquerading as primitive operands What is actually happening here? String s1 = X + Y ; 22 11

12 and while we re at it Let s take another look at Strings masquerading as primitive operands What is actually happening here? String s1 = X + Y ; gets transformed to String s1 = new StringBuffer().append( X ).append( Y ).tostring(); 23 Numeric Strings Using the Wrapper Classes! String s = 1020; int n1 = Integer.parseInt(s); long n2 = Long.parseLong(s); double n2 = Double.parseDouble(s); float n3 = Float.parseFloat(s); number to string conversions? best handled using the + operator! 24 Copyright 12

13 25 How to get primitive values from String objects! suppose we have a sequence of characters! suppose that sequences happens to be the same as a literal value from a primitive type! e.g., l false C!! Use any of these static methods! Integer.parseInt(str)!! Short.parseShort(str)!! Byte.parseByte(str)!! Long.parseLong(str)!! Double.parseDouble(str)!! Float.parseFloat(str)!! Boolean.parseBoolean(str)!! look at API, note the contract re: parameter! java.lang.numberformatexception: Value 25 out of range. How to get primitive values from String objects! suppose we have a one-character String and we want the corresponding char! e.g., C d 9!! there is a wrapper class Character(just like the others)! unfo, there is no Character.parseCharacter(str) or other such static method! instead: char c = C.charAt(0) 26! 26 13

CSE1710. Big Picture. Tuesday, Dec 02 is designated as a study day. No classes. Thursday, Dec 04 is last lecture.

CSE1710. Big Picture. Tuesday, Dec 02 is designated as a study day. No classes. Thursday, Dec 04 is last lecture. CSE1710 Click to edit Master Week text 12, styles Lecture 22 Second level Third level Fourth level Fifth level Fall 2014 Tuesday, Nov 25, 2014 1 Big Picture Tuesday, Dec 02 is designated as a study day.