Following is a list of some of the methods and techniques we encountered in Lesson 3:,,,,, and escape sequences.

Transcription

1 17-1 Following is a list of some of the methods and techniques we encountered in Lesson 3:,,,,, and escape sequences. In Lesson 9 we studied how to compare s: equals( ) and equalsignorecase( ) We will now look at some of the (and examples) of some of the more advanced methods. Recall from Lesson 15 that the layout of a signature is as follows: public returntype methodname( type parameter1, type parameter2, ) The variable and could be,,,,, etc. For the examples below assume that is a as follows: s = The Dukes of Hazzard For convenience, the indices of the individual characters of this are given below: T h e D u k e s o f H a z z a r d Notice this method accepts any Object. Here, we will specifically use a object. The general syntax for usage of is: This method has three rules: a. If alphabetically precedes coat room then it returns a negative. b. If alphabetically comes after coat room it returns a positive. c. If is alphabetically equal to coat room then it returns zero. System.out.println( s.compareto( coat room )); // The reason we get a negative number in the example above is because T alphabetically precedes c. Refer back to Lesson 13 and you will see that the ASCII code for capital T is 84 and the ASCII code for little c is 99. The number 84 comes before 99 so we say that The Dukes of Hazzard comes before (or alphabetically precedes) coat room. There is another version of, that is not case sensitive. This method comes in 6 flavors: (All return 1 if the search is unsuccessful.) Search from left to right for the first occurrence of the. int j = s.indexof( Hazzard ); System.out.println(j); //

2 17-2 Starting at index, search from left to right for the first occurrence of the int j = s.indexof( Hazzard, 15); System.out.println(j); // it couldn t find it when starting at 15 int j = s.indexof( e, 4); System.out.println(j); //. First e is at 2, but we started searching at 4 Search from left to right for the first occurrence of the int j = s.indexof( D ); System.out.println(j); // This method is very similar to c. above, except, instead of a character we give the ASCII code of the character desired. int j = s.indexof(68); // ASCII code for D is 68 System.out.println(j); // Starting at index, search from left to right for the first occurrence of the int j = s.indexof( e, 4); System.out.println(j); // This method is very similar to e. above, except, instead of a character we give the ASCII code of the character desired. int j = s.indexof(101, 4); // ASCII code for e is 101 System.out.println(j); // This method also comes in 6 flavors: (All return 1 if the search is unsuccessful.) These are exactly the same as the method, except here, we begin searching from the right side of the instead of the left as with. Only two examples will be given since they are so similar to the previous examples.

3 int j = s.lastindexof( Haz ); System.out.println(j); // int j = s.lastindexof( Haz, 11); System.out.println(j); // 17-3 This method returns the character at the specified index. char mychar = s.charat(6); System.out.println(myChar); // This method replaces occurrences of the character with the character. String mystring = s.replace( z, L ); System.out.println(myString); // This method replaces occurrences of with. String mystring = s.replace( Dukes, Nerds ); System.out.println(myString); // This method removes whitespace from both ends of the while leaving interior whitespace intact. (Whitespace consists of \n, tab (\t), and spaces.) String s = \t Ding Dong \t \n System.out.println( X + s.trim( ) + X ); // This method returns when this contains the ; otherwise,. boolean b = Sticks and Stones.contains( tic ) This methods returns when this contains as its leading substring. boolean b = Have a good day..startswith( Hav ); // In Lesson 7 we learned how to use the class to input text from the keyboard. Here, we illustrate further uses of in parsing a. Instead of passing to the constructor as we did for keyboard input, we pass a to the constructor as follows:

4 Scanner sc = new Scanner( Please, no more homework! ); 17-4 A delimiter is a series of characters that separates the text presented to a object into separate tokens. The default delimiter is whitespace, so the separate tokens produced by repeated application of the method in the above example would be Please,, no, more, and homework!. The method allows a custom delimiter to be set using regular expressions (see Appendix AC). One of the key concepts here is that of the position of the object as it scans its way through the. We will always think of this position as being two characters in the same way that a cursor in a word processor is never a character; rather it is always two characters (or perhaps preceding the first character, or just after the last character). As tokens are returned by, etc., the position advances to just after the last token returned. Let us consider the following sequence of code where a is passed to the constructor of the class as we illustrate the concept of position. We are also going to introduce the,,, and methods (all use regular expression arguments see Appendix AC). These methods will ; rather, their functions will be : Scanner sc = new Scanner("A string for testing scanner"); //The default delimiter of whitespace will be used. System.out.println(sc.next( )); // System.out.println(sc.findInLine("ri")); // String ns = sc.next( ); //next( ) returns a String System.out.println(ns); // sc.usedelimiter("r\\s+"); // System.out.println(sc.next( )); // sc.skip( r\\s*test ); // System.out.println(sc.next( )); // We should take note of several salient features from this example: 1. The position of a object always moves. It can never be backedup. Likewise, searches (as with e and ) always move from the current position. 2. Starting from the current position and moving forward, the method searches ahead to find the specified substring. Notice above that the ri sought after does have to follow the current position. If the substring is not found, a is returned.

5 In the method the specified substring indeed need to follow the current position. The r test sought after needs to immediately follow the current position. If the substring is not found, a is thrown (an error). In addition to the searching methods and, there is another that can sometimes be useful,. This is identical to except for the additional parameter,, which limits the search for to the next characters. If the search is to be successful, must be found in its within characters. If is zero the search is allowed to continue to the end of the text if necessary. Let s consider what happens when there are two delimiters as illustrated by the following code: String ss = "abcxyxydef"; Scanner sc = new Scanner(ss); sc.usedelimiter("xy"); while(sc.hasnext( )) System.out.println(sc.next( )); The resulting printout is: abc def The blank line is an that was between the two successive delimiters. In order to refine our understanding of the, and, consider how an -spammer might try to disguise a word for the purpose of getting past a spam filter. Instead of transmitting something like Low cost loans, the trick is to send a similar phrase with intervening characters like, L*o*w* *c*o*s*t* lo*a*n*s. Notice this is still readable; however, a standard spam filter would be defeated. In the first line of code below we see an attempt at disguising the word dirty. The remainder of the code strips away the superfluous characters leaving only the original word so that the spam filter can properly detect it. String s = "d^^*_^^ir...-t***y"; //"dirty" Scanner sc = new Scanner(s); sc.usedelimiter(""); //set delimiter to nothing which makes every character a token String answer = "";

6 while(sc.hasnext( )) //skip the stuff we want to get rid of while(sc.hasnext("\\w_")) 17-6 sc.skip("_*"); //skip underscores if(sc.hasnext( )) sc.skip("\\w*"); //skip non-word characters.word characters are[a //za-z_0-9] if(sc.hasnext( )) answer = answer + sc.next( ); System.out.println(answer); //prints In Lesson 7 we learned to input text from the keyboard by creating a object by passing to the constructor. There, we directly used this object to parse the input with,, etc. To avoid some strange effects, it is suggested that all the input be immediately stored into a using and then passed to a object. Then parse this second object using,,, etc. Consider the following program that allows something like to be entered as input from the keyboard. A object then uses the plus signs (and any adjoining whitespace) as delimiters and produces the sum of these numbers(1523). import java.io.*; import java.util.*; public class Tester public static void main(string args[]) Scanner kb = new Scanner(System.in); System.out.print("Enter something like , : "); String s = kb.nextline( ); //Best to store in a String and then create a new Scanner //object; otherwise, it can get stuck waiting for input. Scanner sc = new Scanner(s); //Set delimiters to a plus sign surrounded by any amount of white space...or... // a minus sign surrounded by any amount of white space. sc.usedelimiter("\\s*\\+\\s*"); int sum = 0; while(sc.hasnextint( )) sum = sum + sc.nextint( ); System.out.println("Sum is: " + sum);

7 Enter something like , : , Sum is: 1523 Enter something like , : , Sum is: 1249 Exercise on Lesson 17

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9 Project Encryption/Decryption encryptstring String decryptstring String Tester Enter a sentence that is to be encrypted: This is a very big morning. Original sentence = This is a very big morning. Encrypted sentence = This is a ag',rery dug>?/ijeb..w ssadorninjeb..w. Decrypted sentence = This is a very big morning.