What is printed from this program, please? 01

public static Demo05{
public static void main(String[] args){
for(int i=0; i<3; i++){
for(int j=0; j<6; j++)
if(( i - j > 2 ) || (j > i +1)) System.out.print(' - ');
else System.out.print(' + ');
System.out.pritnln();
Can you also please show the steps of how it is working? I will be greatful
Thanks

If you want to know what's printed, run the program and see for yourself.
If there's something in particular you don't understand, ask about that. But if you just say you don't know what any of it means, then the advice is to start on page 1 of a Java textbook or tutorial.

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My deskjet 2050 will scan from both of my macbook computers, but will only print from one. It gives an "on hold" message whenever I print from this computer. Then I hit "resume" and it says "nothing is being held." I have already downloaded and re-installed the driver, and it didn't make any difference.
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Hello @CazzoInMano , and welcome to the HP Forums, I hope you enjoy your experience!
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After the updates, I would recommend reinstalling the HP Deskjet 2050 All-in-One Printer.
Good luck and please let me know the results of your troubleshooting steps. Thank you for posting on the HP Forums!
Please click “Accept as Solution " if you feel my post solved your issue, it will help others find the solution.
Click the “Kudos, Thumbs Up" on the right to say “Thanks" for helping!
Jamieson
I work on behalf of HP
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I need help instantly on this program please

import java.util.*;
public class D3
          private static int[] z = new int[100000];
private static int first=z[0];
          private static int last=z[n-1];
          private static int n=100000;
public static void main(String args[])
Scanner input=new Scanner(System.in);
for(int i=0;i<z.length;i++)
          z=2*i;
int seqSearch(z;50000;n); //method call 4 key where key=mid
          int binSearch(z;first;last;50000);
          int seqSearch(z;35467;n); //method call 4 key where key in the left half
          int binSearch(z;first;last;35467);
          int seqSearch(z;89703;n); //method call 4 key where key in the right half
          int binSearch(z;first;last;89703);
          public int seqSearch(int z[];int key;int n)
     long start = System.currentTimeMillis();
int count=0;
int ans=-1;
for(int i=0;i<n;i++)
if z[i]=key
count++
{ans=i
break;}
return ans;
long elapsed = System.currentTimeMillis() - start;
System.out.print("Execution Time:" + elapsed);
System.out.print("# of Basic Operations:" + count);
     public int binSearch(int z[];int first;int last;int key)
     long start = System.currentTimeMillis();
     int count=0;
     if(last<first){
     count++;
     index=-1;
     else
     count++;
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     if(ket=z[mid]{
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     else
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// if anyone could tell me whats wrong with my code i'd be greatful...the program is supposed to perform binary and sequential search on a sorted array of 100000 numbers.once on an item in the middle of the array once on the right side of it and once on the left side...i also need to count the number of basic operations for the same number in both sequential and binary to see whats better.and i need to check the time...plz i need help now,,,

"Guide to a first-time poster"
you need to add exclamation marks to signify how urgent it is
e.g.
i need help instantly on this program please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
capital letters is better
I NEED HELP INSTANTLY ON THIS PROGRAM PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
starting the italics on line 1, better again
import java.util.*;
public class D3
private static int[] z = new int[100000];
private static int first=z[0];
private static int last=z[n-1];
private static int n=100000;
public static void main(String args[])
Scanner input=new Scanner(System.in);
for(int i=0;i<z.length;i++)
z=2*i;
int seqSearch(z;50000;n); //method call 4 key where key=mid
int binSearch(z;first;last;50000);
int seqSearch(z;35467;n); //method call 4 key where key in the left half
int binSearch(z;first;last;35467);
int seqSearch(z;89703;n); //method call 4 key where key in the right half
int binSearch(z;first;last;89703);
public int seqSearch(int z[];int key;int n)
long start = System.currentTimeMillis();
int count=0;
int ans=-1;
for(int i=0;i<n;i++)
if z=key
count++
{ans=i
break;}
return ans;
long elapsed = System.currentTimeMillis() - start;
System.out.print("Execution Time:" + elapsed);
System.out.print("# of Basic Operations:" + count);
public int binSearch(int z[];int first;int last;int key)
long start = System.currentTimeMillis();
int count=0;
if(last><first){
count++;
index=-1;
else
count++;
int mid=(first+last)/2
if(ket=z[mid]{
index=mid;
else
if(key><z[mid]){
index = binSearch(z[];first;mid-1;key);
else
index=binSearch(z[];mid+1;last;key);
return index;
long elapsed = System.currentTimeMillis() - start;
System.out.print("Execution Time:" + elapsed);
System.out.print("# of Basic Operations:" + count);
and what about the dukes, offer 10 (never to be awarded, of course)
do this, then sit back and watch the replies roll in.

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We couldn't print PDF's prior to the replacment of the printer, but I thought replacing the printer might help.
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I'm sorry to bother you with such stupid questions. I have to admit I'm just making my first steps in programming.
So I have a program like this and don't understand how it works:
// Masking sign extension.
class HexByte {
static public void main(String args[]) {
char hex[] = {
'0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'a', 'b', 'c', 'd', 'e', 'f'
byte b = (byte) 0xf1;
System.out.println("b = 0x" + hex[(b >> 4) & 0x0f] + hex[b & 0x0f]);
Specifically, how could 0xf1 be represented in binary? As far as I know 11110001 contains negative value. Moreover 0xf1 in 10's base is 241, so how can I convert it to byte? Doin so, I'm gonna lose its value (since the byte range is from -128 to 127). And why do I need it? I even have no idea how the variable would look like in binary after converting to byte, not mentioning last part of this program.
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AntShay wrote:
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And by the way, is there any method to convert the value of byte to a binary string? I only figured that it might work this way: Integer.toBinaryString(b&0xff), but is there Byte.toBinaryString method or something?It is always okay to up cast, as sign extension will convert correctly. If you don't like the long integer output, you may do a substring() on the output. It is also not hard to roll your own.

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import java.util.*;         //supplies random class
import java.awt.*;          //supplies user interface classes
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private int[] values;
private int size;
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Message Edited by Marc A on 01-12-2007 11:03 AM
Attachments:
Word_print.PNG ‏18 KB

goldn:
Welcome to the Apple Discussions. Just ran a test and I got a perfect card. Here's what I did.
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3 - click on the Advanced button
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10 - click on Print
11 - when the paper exits the printer quickly put it back in the printer
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<Edited by Host>

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What is printed from this program, please? 01

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