saipavan vallabhaneni

thanks himanshu and abhi

i understood now

sorry Himanshu,
i didn't understand what you were trying to tell

1.4E-45 -1.4E-45
-2147483648 -2147483648
-1 1
integer max value: 2147483647

Output completed (0 sec consumed)

i got the following output...
int m = Integer.MIN_VALUE; //-2147483648 int n = -m; //2147483648 System.out.println( m+ " "+n);

why is n being assigned a -ve value where we assigned -m to n

public class TestClass { public static void main(String[] args) throws Exception { float a = Float.MIN_VALUE; float b = -a; System.out.println( a+ " "+b); int m = Integer.MIN_VALUE; int n = -m; System.out.println( m+ " "+n); int x = -1; int y = -x; System.out.println(x +" " + y); } }

why are the m and n values same??? though we have assigned a -m to n...

thanks ankit and punit

new Person("sai")..the hashCode returns 3 and so in the corresponding bucket using equals method search for "sai" should be made and the corresponding value "11" must b retrieved ..but i am not getting it

import java.util.*; public class Person { private String name; public Person(String name) { this.name = name; } public boolean equals(Person p) { return name.equals(p.name); } public int hashCode() { return name.length(); } public static void main(String args[]) { Map st = new HashMap(); Person p1 = new Person("sai"); Person p2 = new Person("pavan"); st.put(p1,"11"); st.put(p2,"22"); System.out.println(st.get(new Person("sai"))); } }// im getting the output as null instead of 11

though i had over ridden the hashCode() and equals() method i am unable to retrieve the stored value... y is it soo?

hi Emanuel,
the following code might help you

import java.util.*; import java.util.regex.*; import java.io.*; public class ScannerOnFile { public static void main(String[] args) { try{ String a; //Pattern p = Pattern.compile(".\\d\\d"); String p = "\\d\\d"; File file=new File("archivo.txt"); file.createNewFile(); /*FileReader fr=new FileReader(file); BufferedReader br=new BufferedReader(fr); while((a=br.readLine())!=null){ System.out.println(new Scanner(a).findInLine(p)); }*/ Scanner s=new Scanner(file); while (s.findInLine(p) != null) { System.out.println(s.next()); } }catch(Exception e){ System.out.println("error"); } } }

a scanner can directly take file as an argument...and the scanner's findInLine() can take only a String...so specify the pattern you want in a String and pass it to findInLine() and check the result

thanks Punit,
thanks for the detailed explanation

[LIST][*]why is the 1st method invokation getting comppiled even when the types are different
[*]why is showing error for 4th invokation to which the values passed are of same type and result is being casted to int ???

import java.util.*; class WrapperCheck { public <T extends Comparable> T findLarger(T x, T y) { if(x.compareTo(y) > 0) { return x; } else { return y; } } public static void main(String[] args) { WrapperCheck t = new WrapperCheck(); Object z = t.findLarger(123, "456"); //y is this getting compiled??? when the parameters being passed are different types //int m = t.findLarger(123, new Double(456)); compiler shows error //int y = t.findLarger(123, new Integer(456)); compiler shows error //int x = (int) t.findLarger(new Double(123), new Double(456)); what's wrong with this??? we are passing the same type of args to this ...y is it askin req int // found Double } }

thanks Himanshu

when i compile the program below its giving an error...i am unable to to understand whats wrong with the program
error::::::::CollectionReverse.java:11: foreach not applicable to expression type
for (Object obj : reverse(list))

import java.util.*; class CollectionReverse { public static Iterator reverse(List list) { Collections.reverse(list); return list.iterator(); } public static void main(String[] args) { List list = new ArrayList(); list.add("1"); list.add("2"); list.add("3"); for (Object obj : reverse(list)) System.out.print(obj + ","); } }

thanks Ankit,
i am now a little aware of working of the mehods ...

thanks henry,
but in the 1st code snippet having source "ab34ef" why is the group() method returning a null(when start() returns 0)instead of "a"(since it has 0 or more digits)...group() method returns the match that has been found(which in my guess is "a" rather than null as returned by the group())

thanks ankit,

but is it not true that greedy quantifier looks at the entire source string once and then reverts back from right to find the match and include the part of the source left side to the match as the final match...

source: yyxxxyxx
regex: .*xx
output: yyxxxyxx(at match is found at the end and part source string prior to the match is included in the output as the entire source ends in a xx)