EvilZone
Programming and Scripting => Java => : Deque November 13, 2013, 08:54:36 AM
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For this challenge series I will ask some Java questions. The first one to answer the question correctly is the winner. Once the question is answered the winner is free to post the next quiz him-/herself, or -- if he/she has no idea for a challenge I will post the next quiz.
First quiz:
int i = (byte) + (char) - (int) + (long) - 1;
System.out.println(i);
Does this compile?
Explain what is happening and why.
Quizlers:
Neea: 1
Snayler: 1.5
ca0s: 1
Mordred: 1/2
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Love the idea for the quiz :) Makes me want to learn java better.
Yes it compiles and the result is 1
You're basically mixing casts with operators ....
So you first have -1, then you cast it to long so it's gonna be long -1
Then you have + unary operation so it remains long -1, then you cast it to int => int -1
Then you have - operation => int 1, after cast to char => char 1
After + operation so it's gonna be char 1
Then cast to byte => byte 1
And in the end implicitly cast to int => int 1
Think that's about right.
I have no idea for the next challenge cuzz i don't know java that well :)
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Love the idea for the quiz :) Makes me want to learn java better.
Yes it compiles and the result is 1
You're basically mixing casts with operators ....
So you first have -1, then you cast it to long so it's gonna be long -1
Then you have + unary operation so it remains long -1, then you cast it to int => int -1
Then you have - operation => int 1, after cast to char => char 1
After + operation so it's gonna be char 1
Then cast to byte => byte 1
And in the end implicitly cast to int => int 1
Think that's about right.
I have no idea for the next challenge cuzz i don't know java that well :)
Are you sshackss?
If so I would like you to answer only in one forum and leave the fun for others.
Your answer is right. Next quiz:
public class Foo {
public static System sys = null;
public static void main(String[] args) {
sys.out.println("Foo");
}
}
Does that work?
Why or why not?
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"sys" object is null, such an exception can cause a black hole to appear. Avoid null objects at all costs!
BAM!
My question:
public class Foo {
public static void main(String[] args) {
System.out.println("Line1"+
"Line2"+
"Line3"+
"Java is the shit!!");
}
}
Does it print each line in a new line.
Also fuck CODE formatting...
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I'll start by saying I've never touched on Java.
No. It will print out:
Line1Line2Line3Java is the shit!!
"\n" is the shit!!
Oh, and I have no idea about the next challenge..
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"sys" object is null, such an exception can cause a black hole to appear. Avoid null objects at all costs!
BAM!
Your answer is wrong. You get -1 for not waiting for confirmation.
My quiz is still up.
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Ok, my bad. I can't understand why that code does what it does, I mean, there's a null in there, for christ sake!
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out is a static member of the System class, so that code doesn't really use the sys object, but the class instead?
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My answer:
It will work, as in it will display "Foo" on the screen (no compilation error).
Although the definition's correct as far as syntax goes, the Foo class tries to instantiate a variable called sys as a System object (albeit as null).
Unfortunately, although System, as all other classes in Java, extends Object, it is defined in the API as:
public final class System extends Object
So it is a final class, which means it cannot be instantiated at all. All calls to the System class are made directly, through a syntax such as:
System.out.println("Foo");
System.out.printf("Foo%n");
That null tho has me a bit worried, but I think this should be the correct answer :) .
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:o I am surrounded by people way smarter than me...
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My answer:
It will work, as in it will display "Foo" on the screen (no compilation error).
Although the definition's correct as far as syntax goes, the Foo class tries to instantiate a variable called sys as a System object (albeit as null).
Unfortunately, although System, as all other classes in Java, extends Object, it is defined in the API as:
public final class System extends Object
So it is a final class, which means it cannot be instantiated at all. All calls to the System class are made directly, through a syntax such as:
System.out.println("Foo");
System.out.printf("Foo%n");
That null tho has me a bit worried, but I think this should be the correct answer :) .
Final means that it cannot be subclassed. A final class can be instantiated.
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out is a static member of the System class, so that code doesn't really use the sys object, but the class instead?
You nailed it.
Do you have another quiz?
Foo class tries to instantiate a variable called sys as a System object (albeit as null).
sys is not instantiated, there is no object created from the System class in that code. ;)
And, yes, as ca0s said correctly it has nothing to do with System being final.
System can indeed not be instantiated, but the reason for that is the private constructor.
Let's look at the code of System:
/** Don't let anyone instantiate this class */
private System() {
}
public final static PrintStream out = null;
As you can see out is a static member of the class, which is the reason you access it usually via System.out
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Oh true. Welp, I guess they should've taught us this in one of the 5 Java courses I took during University ^^ (or should I say: "they probably taught us but I didn't give enough of a fuck"? ::) )
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Do you have another quiz?
Sorry, I don't. I'm not that fluent in Java to be able to propose an interesting quiz.
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Alright. The next one is an (hopefully) easy question.
As you know Java has the access modifiers: private, public and protected.
What is the default access level if you don't use an access modifier?
Example:
public class AccessTest {
void methodWithoutModifier() { }
}
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The Default access level means that it can be accessed only from within the same package. That is different from public, which can be accessed from outside the package as well.
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The Default access level means that it can be accessed only from within the same package. That is different from public, which can be accessed from outside the package as well.
How is that access level called?
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How is that access level called?
protected
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How is that access level called?
package-private
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package-private
Correct.
Well, now Mordred and you gave one half of the solution each. So one of you is free to make the next quiz. (Or I do it)
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noooo, I didn't make it in time to answer with the name :(
But maybe I have one:
While testing some code that you are developing, you notice that an ArrayIndexOutOfBoundsException is thrown. What is the appropriate reaction?
1. Enclose the offending code in a try block, with a catch block for ArrayIndexOutOfBoundsException that does nothing.
2. Enclose the offending code in a try block, with a catch block for ArrayIndexOutOfBoundsException that prints out a descriptive message.
3. Declare that the method that contains the offending code throws ArrayIndexOutOfBoundsException.
4. None of the above.
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It is fallacious to blindly follow some sort of archetypal convention in the face of a given scenario, especially with something as general and out of context as an exception.
An ArrayIndexOutOfBoundsException could be thrown as a result of poor and/or rudimentary argument parsing (assigning variables to args[n]) and not specifying an option when running the program.
Depending on your situation, for the most part if you're intending on building an application that will be employed by end users, you'd likely pick 2). For personal projects and developer-oriented software, you could do either 1) or 2), most likely you'd do an e.printStackTrace().
In the face of ambiguity, I pick "none of the above".
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1) is a terrible suggestion as you just ignore the problem and make it disappear from the eye of the programmer.
2) won't solve anything either, but at least you still see the problem.
3) is completely useless. ArrayIndexOutOfBoundsException is an unchecked exception. The compiler doesn't force you to handle it. It won't make any difference if you write throws in the method definition.
An ArrayIndexOutOfBounds is in almost all cases the fault of the programmer.
The answer would be 4, because you have to correct the code instead of handling an exception somewhere.
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Indeed, the correct answer is 4.
That is an actual exam-question also. I took it from my OOP2 exam from my 2nd year in Uni. The answer that we were supposed to give (as told to us by our lecturer) is:
None of the above. Because ArrayIndexOutOfBounds is an RunTimeerror and it should be fixed by the coder.
[offtopic]
The teacher was terrible by the way. Mostly had to learn Java on my own by practice. I only managed to pass by doing 90% of the shit alone after he gave us a list of what we were gonna study in a given period.
@vezzy: I gave a similar answer in which I complained about the ambiguity, but I got 0 points for it...
[/offtopic]
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Alright, next quiz. This is a nice one, someone else asked in another forum:
Why the hell does this compile?
class TestURL {
public static void main(String... args) {
http://www.hackcommunity.com
System.out.println("Why don't you show any error?");
}
}
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This is a remnant from Java's origins in C/C++, and more generally Assembly.
It interprets the "http:" as a section. A section that can then be arbitrarily used and jumped to for various purposes. The two trailing slashes then enclose the remaining part (the global subdomain+hostname) as a comment, so it does not get executed and thus does not cause any errors.
To illustrate a common usage for sections with C:
#include <stdio.h>
int main(void) {
int x = 5;
if (x <= 5) {
goto error;
}
}
error:
fprintf(stderr, "Operation failed.");
Consequently, let us restructure your example in Java to see more clearly what is actually being done:
class TestURL {
public static void main(String... args) {
http:
// www.hackcommunity.com
System.out.println("Why don't you show any error?");
}
}
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This is a remnant from Java's origins in C/C++, and more generally Assembly.
It interprets the "http:" as a section. A section that can then be arbitrarily used and jumped to for various purposes. The two trailing slashes then enclose the remaining part (the global subdomain+hostname) as a comment, so it does not get executed and thus does not cause any errors.
To illustrate a common usage for sections with C:
#include <stdio.h>
int main(void) {
int x = 5;
if (x <= 5) {
goto error;
}
}
error:
fprintf(stderr, "Operation failed.");
Consequently, let us restructure your example in Java to see more clearly what is actually being done:
class TestURL {
public static void main(String... args) {
http:
// www.hackcommunity.com
System.out.println("Why don't you show any error?");
}
}
It's called a label and can only be jumped to via break or condition statement.
You got it, you can make the next quiz.
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public class Example {
public static void main(String[] args) {
int x = 8;
double y = 4.671268942105652;
x = x + y;
System.out.println(x);
}
}
Will this compile? If so, why? If not, why and how would you fix it?
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It will not compile, because you cannot convert from double to int without data loss.
How would I fix it? Depends, if i want a floating point result, chance type of x to double, if i want an int result add cast to int and deal with data loss.
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Looking from a science standpoint, you would go by significant figures, which would be 8 in this case. So I would cast int.
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Both sufficient, but with particular notice to Neea. You'd especially want to go by significant figures if in finance.
Take the next one, Neea.
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Say you have the following classes
public abstract class Symbol {
private char symbol;
......... //constructor
@Override
public String toString() {
return new Character(symbol).toString();
}
}
public class Nonterminal extends Symbol {
........ //constructors
@Override
public String toString() {
return super.toString();
}
}
public class Terminal extends Symbol {
..... //constructors
@Override
public String toString() {
return super.toString();
}
}
Will this compile or not, will it print out "it works" or not and why, and how would you make it work?
......
Nonterminal nt = new Nonterminal("S");
Terminal st = new Terminal("S");
if(nt.equals(st)){
System.out.println("It works");
}
.....
Not a particularly hard question, but oh well .... my java knowledge is fairly limited.
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Say you have the following classes
public abstract class Symbol {
private char symbol;
......... //constructor
@Override
public String toString() {
return new Character(symbol).toString();
}
}
public class Nonterminal extends Symbol {
........ //constructors
@Override
public String toString() {
return super.toString();
}
}
public class Terminal extends Symbol {
..... //constructors
@Override
public String toString() {
return super.toString();
}
}
Will this compile or not, will it print out "it works" or not and why, and how would you make it work?
......
Nonterminal nt = new Nonterminal("S");
Terminal st = new Terminal("S");
if(nt.equals(st)){
System.out.println("It works");
}
.....
Not a particularly hard question, but oh well .... my java knowledge is fairly limited.
No one here has been trying to answer it by now, so I will to keep this series going.
It will compile (assumed that the code parts not given are correct).
It won't print "it work's". The reason is the default implementation of equals:
public boolean equals(Object obj) {
return (this == obj);
}
The == will only be true if the memory addresses of the objects are the same. In this case they aren't.
In order to make it work you would have to override equals in Symbol.
Example implementation:
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (obj instanceof Symbol) {
Symbol other = (Symbol) obj;
if (symbol != other.symbol)
return false;
}
return true;
}
//if you change equals you have to change hashCode too
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + symbol;
return result;
}
But: It is a bad idea to make objects with two different subtypes equal to each other. They wouldn't need different types if they where equal.
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Your answer is correct, but if you need to compare the 2 values of the different objects, which was the case in my example you don't need to override the equals method, just use toString() and compare both as string objects. This was the reason why i let the toString() methods visible in my example. Please continue :)
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Your answer is correct, but if you need to compare the 2 values of the different objects, which was the case in my example you don't need to override the equals method, just use toString() and compare both as string objects. This was the reason why i let the toString() methods visible in my example. Please continue :)
Ah, I didn't know that I can touch this part of the code, because then you could just do
System.out.println("It works");
And it is alright too.
However, next quiz:
... x = ...
if (x != x) System.out.println("Hello Evilzone");
What do you have to put into the ... to make it print out "Hello Evilzone"?
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double x = Double.NaN;
if (x != x) System.out.println("Hello Evilzone");
you use a NaN (not a number)which is an undefined value. NaN can not equal NaN because many different operations can result in NaN such as 0 divided by 0,or log(-1).
the double class has a built-in static NaN value use here.
i can't think of anything to use for a quiz at the moment.
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double x = Double.NaN;
if (x != x) System.out.println("Hello Evilzone");
you use a NaN (not a number)which is an undefined value. NaN can not equal NaN because many different operations can result in NaN such as 0 divided by 0,or log(-1).
the double class has a built-in static NaN value use here.
i can't think of anything to use for a quiz at the moment.
Correct
Next quiz:
long l = 1000000000 + 2000000000;
System.out.println(l);
What is the value of l and why?
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3.000.000.000, because l is of type long, which can support numbers from -2^63 to (2^63)-1 [signed].
If it was an int, it couldn't possibly hold the end result of the operation since an int can hold a maximum value of around 2 billion.
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3.000.000.000, because l is of type long, which can support numbers from -2^63 to (2^63)-1 [signed].
If it was an int, it couldn't possibly hold the end result of the operation since an int can hold a maximum value of around 2 billion.
Run it and reconsider your answer. ;)
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it is definitely strange. it being cast as an int for some reason. :-\
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Next quiz:
long l = 1000000000 + 2000000000;
System.out.println(l);
What is the value of l and why?
The value of "l" is -1294967296.
This happens because the arithmetic operation is being done with two int numbers. Assigning the result to a long variable is irrelevant. In this case, the answer would be out of bound for an int, so it returns this strange value.
The fix:
long l = 1000000000;
l += 2000000000;
System.out.println(l);
I have no quiz
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The value of "l" is -1294967296.
This happens because the arithmetic operation is being done with two int numbers. Assigning the result to a long variable is irrelevant. In this case, the answer would be out of bound for an int, so it returns this strange value.
The fix:
long l = 1000000000;
l += 2000000000;
System.out.println(l);
I have no quiz
Correct.
An alternative fix:
long l = 1000000000 + 2000000000L;
System.out.println(l);
Next quiz:
double c = 0.1;
if (c * 0.1 == 0.01) {
System.out.println("Hell yeah.");
} else {
System.out.println("Nooooo.");
}
What does it print and why?
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Next quiz:
double c = 0.1;
if (c * 0.1 == 0.01) {
System.out.println("Hell yeah.");
} else {
System.out.println("Nooooo.");
}
What does it print and why?
It would print "Nooooo.".
The equation would result in a very large amount of zeroes, and most calculators or other things would just round it up to 0.01, however because this is a double it goes the smallest a double can go, being
.010000000000000002
At least this is my interpretation of it, I may be wrong.
If I'm correct, I don't have a quiz.
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It would print "Nooooo.".
The equation would result in a very large amount of zeroes, and most calculators or other things would just round it up to 0.01, however because this is a double it goes the smallest a double can go, being
.010000000000000002
At least this is my interpretation of it, I may be wrong.
If I'm correct, I don't have a quiz.
The explanation is not correct.
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The explanation is not correct.
Ah well I did my research this time.
A double means double-precision, containing 64 bits while a float contains 32.
1/10 can't be represented by a binary fraction so the computer scales the answer down to the closest fraction it can get legitimately.
If I'm incorrect again would you mind shooting me a pm containing the correct explanation? Thanks.
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Ah well I did my research this time.
A double means double-precision, containing 64 bits while a float contains 32.
1/10 can't be represented by a binary fraction so the computer scales the answer down to the closest fraction it can get legitimately.
If I'm incorrect again would you mind shooting me a pm containing the correct explanation? Thanks.
That one is correct. I will post the next quiz tomorrow or someone else who has one can post it too. Gtg
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Is the following a valid statement? Why?
String [] foo [] = null;
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I think it is valid. It is two dimensional String.
String[][] foo && String[] foo [] && String foo [][]
all are the same..
I don't have any new quiz if I'm right.
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Correct.
public class Home {
private static Home home = new Home();
private static int DEFAULT_NR_OF_ROOMS = 4;
private final int rooms;
public Home() {
rooms = DEFAULT_NR_OF_ROOMS - 1;
}
public static void main(String[] args) {
System.out.println(home.rooms);
}
}
What does this print and why?
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Correct.
public class Home {
private static Home home = new Home();
private static int DEFAULT_NR_OF_ROOMS = 4;
private final int rooms;
public Home() {
rooms = DEFAULT_NR_OF_ROOMS - 1;
}
public static void main(String[] args) {
System.out.println(home.rooms);
}
}
What does this print and why?
Well the problem is with int rooms because it is final and at declaration it is not assigned any value. Final variables must be assigned values at their declaration.
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Well the problem is with int rooms because it is final and at declaration it is not assigned any value. Final variables must be assigned values at their declaration.
Nope. It is totally fine to initialize final fields in the constructor and if you don't init final fields properly the compiler will complain. But this code compiles and runs without a problem.
The question is: Why this output?
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It outputs 3.
The home member object was used inside the class so access specifiers are irrelevant.
fun fact: each home object has a static home object within it that can be referenced infinitely so this is also a legal statement
System.out.println(home.home.home.home.home.rooms); //homes inside of homes inside of homes. its Homeception
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It outputs 3.
The home member object was used inside the class so access specifiers are irrelevant.
Wrong.
Run the code and reconsider your answer.
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Wrong.
Run the code and reconsider your answer.
Well, it still outputs 3. was that part at least right?
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Well, it still outputs 3. was that part at least right?
It does not output 3. What VM are you using?
Edit: Ok, I overlooked a part in the JVM specification about final fields (I didn't know that putting in final would change it and the code I put here first was working the way I wanted to). Try now with the code above.
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It outputs -1
You are calling constructor before you assign the value to the static variable. So in the constructor code, there no value of static variable.
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It outputs -1
You are calling constructor before you assign the value to the static variable. So in the constructor code, there no value of static variable.
Correct. Reason can be found here at step 9
http://docs.oracle.com/javase/specs/jls/se5.0/html/execution.html#12.4.2
It also explains why it didn't work with final. Final fields are initialized before the others.
You can make the next quiz.
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Next quiz, same question. What will be printed and why?
import java.util.HashSet;
import java.util.Set;
public class Name {
private final String first, last;
public Name(String first, String last) {
this.first = first;
this.last = last;
}
@Override
public boolean equals(Object o) {
if (!(o instanceof Name))
return false;
Name n = (Name) o;
return n.first.equals(first) && n.last.equals(last);
}
public static void main(String[] args) {
Set<Name> s = new HashSet<Name>();
s.add(new Name("Mickey", "Mouse"));
System.out.println(s.contains(new Name("Mickey", "Mouse")));
}
}
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It returns false.
Because we're using an instance of HashSet<> as Set; and this HashSet uses the methods hashCode() and equals() to check if two items given to the Set are equal or not.
Since we haven't implemented an extra hashCode method, it uses the standart hashCode() from Object and there we will get two different codes for those two objects.
// Edit: added the explanation
Example to get true:
import java.util.HashSet;
import java.util.Set;
public class Name
{
private final String first, last;
public Name( String first, String last )
{
this.first = first;
this.last = last;
}
@Override
public boolean equals( Object o )
{
if ( !( o instanceof Name ) )
return false;
Name n = (Name) o;
return n.first.equals( first ) && n.last.equals( last );
}
@Override
public int hashCode()
{
return( first.hashCode() + last.hashCode() ); // this isn't a good hashCode method! Don't do it in real but here it works
}
public static void main( String[] args )
{
Set<Name> s = new HashSet<Name>();
s.add( new Name( "Mickey", "Mouse" ) );
System.out.println( s.contains( new Name( "Mickey", "Mouse" ) ) );
}
}
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It returns false.
Because we're using an instance of HashSet<> as Set; and this HashSet uses the methods hashCode() and equals() to check if two items given to the Set are equal or not.
Since we haven't implemented an extra hashCode method, it uses the standart hashCode() from Object and there we will get two different codes for those two objects.
// Edit: added the explanation
Example to get true:
import java.util.HashSet;
import java.util.Set;
public class Name
{
private final String first, last;
public Name( String first, String last )
{
this.first = first;
this.last = last;
}
@Override
public boolean equals( Object o )
{
if ( !( o instanceof Name ) )
return false;
Name n = (Name) o;
return n.first.equals( first ) && n.last.equals( last );
}
@Override
public int hashCode()
{
return( first.hashCode() + last.hashCode() ); // this isn't a good hashCode method! Don't do it in real but here it works
}
public static void main( String[] args )
{
Set<Name> s = new HashSet<Name>();
s.add( new Name( "Mickey", "Mouse" ) );
System.out.println( s.contains( new Name( "Mickey", "Mouse" ) ) );
}
}
Correct. It's your turn now. Post a quiz.
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class Quiz
{
public static void main( String args[] )
{
int x = 0;
int y = 55;
try
{
System.out.println( y/x );
}
catch( Exception e )
{
e.printStackTrace();
}
}
}
Will there be an exception thrown?
If yes:
when will it be thrown?
At compiletime? or
Runtime?
what Exceptiontype is it?
Why is it thrown at this specific time ( Runtime / Compiletime ) ?
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An exception will be thrown because you are dividing by zero, at run-time because this is a semantic, not syntax error, and the exception is "ArithmeticException"
If im correct I dont have the next quiz.
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It's correct.
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System.out.println(0610 + 0b10);
What is the result of that addition and why?
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The result is 394, because 0610 ( = 392|10 ) is a number written in octal number system, which you can see on the leading 0 and 0b10 ( = 2|10 ) is one in binary system, because of the leading 0b. This binary notation works since SE 7.
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The result is 394, because 0610 ( = 392|10 ) is a number written in octal number system, which you can see on the leading 0 and 0b10 ( = 2|10 ) is one in binary system, because of the leading 0b. This binary notation works since SE 7.
Correct. Do you have a next quiz?
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Yep, I have one:
class Clazz extends Runnable
{
private volatile boolean stop = false;
public void stop()
{
stop = true;
}
@Override
public void run()
{
while( !stop )
{
// do Something
}
}
}
Why does the variable stop have to be volatile?
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I would like to carry on with the quiz and since no one answered the last question I will try.
volatile is there to prevent optimizations of the compiler that might interfere with threaded access to the field; especially caching is a problem here.
So if you create a thread with that Runnable, the volatile makes sure you will always get the right value.
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Yep, thats right!
But, I talked to my lecturer today if I have to use it always or not, because of the leaking/ caching problem . He said, he never used it and also didn't have problems, so maybe it's already fixed by the JVM. If someone does have more information about it, it would be nice if he/she would share it with me.
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Yep, thats right!
But, I talked to my lecturer today if I have to use it always or not, because of the leaking/ caching problem . He said, he never used it and also didn't have problems, so maybe it's already fixed by the JVM. If someone does have more information about it, it would be nice if he/she would share it with me.
Fixed?
This is nothing to be fixed. Caching is done to optimize, to get more speed. This is on purpose. If your professor says such things, this is just a big facepalm.
And of course you have to use volatile in certain circumstances if you use threads. Just because your professor didn't get a problem, it doesn't mean that a problem is not possible with code that doesn't use volatile where it should be used. Especially threaded code is hard to test, the order of execution of the threads might be different every time you run the code. So it is a bit of luck to get the problematic execution orders while testing.
See here for more explanation: https://stackoverflow.com/questions/106591/do-you-ever-use-the-volatile-keyword-in-java (https://stackoverflow.com/questions/106591/do-you-ever-use-the-volatile-keyword-in-java)
Next quiz:
public static void main(String... args) {
float i = 0;
while (i != i + 1) {
i++;
}
System.out.println(i);
}
Is this an endless loop?
Why (or why not)?
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Next quiz:
public static void main(String... args) {
float i = 0;
while (i != i + 1) {
i++;
}
System.out.println(i);
}
Is this an endless loop?
Why (or why not)?
I believe it is an endless loop, 'i' can never be higher than 'i'+1,
which means that the check 'i' != 'i'+1 is always True.
I have no clue about Java so feel free to create a new quiz.
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I think I inartfully expressed what I mean.
I meant it like, the JVM maybe already recognises that there's a variable used in a multithread system. So it's updating it in a way you won't get problems because of the caching (like MESI does on a physical multicore system). But yep, the reordering-problem still exists.
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I believe it is an endless loop, 'i' can never be higher than 'i'+1,
which means that the check 'i' != 'i'+1 is always True.
Nope.
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I dont really understand everything, but i think it has something to do with the finite precision of floating point numbers. There is a certain point where x == x + 1 because of this. For these same reasons you have to be careful of comparing and doing important calculations with them because of their imprecision.
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I dont really understand everything, but i think it has something to do with the finite precision of floating point numbers. There is a certain point where x == x + 1 because of this. For these same reasons you have to be careful of comparing and doing important calculations with them because of their imprecision.
That is correct.
float has 6 to 9 significant decimal digits precision. These are represented in the mantissa which uses 23 bits. Additionally there is 1 sign bit and 8 bits for the exponent.
Now imagine you have a pretty high number, the bits in the mantissa won't be enough to hold all of the digits. The exponent part makes it still possible to use very high numbers (or small ones), but the distance between two consecutive numbers that are representable with float grows with the numbers growing. So once you get to the point where this distance is higher than 1, (x == x + 1) is true, because the 1 you add will just be ignored as (x+1) is not representable anymore.
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I apologize. I forgot to say that I dont have the next quiz if it's correct. Great question, by the way, Deque. Im learning a lot from these :)
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What happens here?
for (byte b = Byte.MIN_VALUE; b < Byte.MAX_VALUE; b++) {
if (b == 0x90) {
System.out.print("Joy!");
}
}
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You're looping for the minimum and maximum value of a byte as defined by Java's Byte class and printing text in the event of an Intel NOP instruction. This does not occur and hence nothing is printed.
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You're looping for the minimum and maximum value of a byte as defined by Java's Byte class and printing text in the event of an Intel NOP instruction. This does not occur and hence nothing is printed.
Yeah, but why does it not occur?
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Hmmm.
Well, 0x90 in decimal is equal to 144, so that's outside of the byte range (up to 127).
Thus I'm assuming this is an actually integer, and to get this to work you'll either need to cast 0x90 to byte or convert the byte to an integer.
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Hmmm.
Well, 0x90 in decimal is equal to 144, so that's outside of the byte range (up to 127).
Thus I'm assuming this is an actually integer, and to get this to work you'll either need to cast 0x90 to byte or convert the byte to an integer.
Your explanation is correct, do you have the next quiz?
(Sorry for editing your post. I pressed the wrong button while trying to quote you, but I didn't change anything)
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Pass the quiz on to the next person.
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The next one is not a program, but a question.
What is the magic number of a .class file and where did it come from?
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The magic number is a hex number meant to identify the file format. It is 0xCAFEBABE.
Came from a combination of eating at a cafe and a band who used to play there (grateful dead) and a friend dying there. Not sure whether it was the band or the actual death but then the founder of java and friends started calling the place cafe dead. james gosling realized CAFEDEAD was a hex number and started using it for the persistent object format. Deciding on cafe as a theme, he grepped for characters that made sense with it and got CAFEBABE.
Dont have next quiz if im correct, and nice question :)
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The magic number is a hex number meant to identify the file format. It is 0xCAFEBABE.
Came from a combination of eating at a cafe and a band who used to play there (grateful dead) and a friend dying there. Not sure whether it was the band or the actual death but then the founder of java and friends started calling the place cafe dead. james gosling realized CAFEDEAD was a hex number and started using it for the persistent object format. Deciding on cafe as a theme, he grepped for characters that made sense with it and got CAFEBABE.
Dont have next quiz if im correct, and nice question :)
You are correct.
Does anybody else have a quiz? You are free to go.
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Sure :)
Let's say you have the following class Man:
public class Man implements java.io.Serializable {
public String name;
public String number;
public transient int age;
public boolean male;
}
And the following PersonInLog class:
public class PersonInLog {
public static void main(String[] args) {
Man e = new Man();
e.name = "Bill";
e.number = "555-555-5555";
e.age = 30;
e.male = true;
try {
FileOutputStream personFile = new FileOutputStream("/tmp/bill.ser");
ObjectOutputStream out = new ObjectOutputStream(personFile);
out.writeObject(e);
out.close();
personFile.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
And finally this:
import java.io.*;
public class PersonOutLog {
public static void main(String[] args) {
Man e = null;
try {
FileInputStream personFile = new FileInputStream("/tmp/bill.ser");
ObjectInputStream in = new ObjectInputStream(personFile);
e = (Man) in.readObject();
in.close();
personFile.close();
} catch (IOException e) {
return;
} catch (ClassNotFoundException c) {
return;
}
System.out.println("Name: " + e.name);
System.out.println("Number: " + e.number);
System.out.println("Age: " + e.age);
System.out.println("Male: " + e.male);
}
}
What will be printed?
Edit:
100 posts!
Also, just to clear it up this is a trick question in a way. Just throwin that out there.
Edit 2:
Excuse me for all the mistakes this is pseudo code written while half asleep. I think I've fixed most of them by now though.
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Name: Bill
Number: 555-555-5555
Age: 30
Male: true
If you were to attempt to read the Man object using your PersonOutLog class, then the above would be printed.
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Excuse me I forgot I posted this, I was going to revise it the next day..
This is a completely unfair question, I don't tell you what class I'm running or in which order. If I said you run PersonInLog and the PersonOutLog, then yes you would be correct.
Forgive me, I wasn't thinking when I wrote this. Consider Lac correct :)
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:) Yeah that's why I made sure to say IF you try to call the Man object.
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So what's the next question ? Or has the previous question been accepted as complete.
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So what's the next question ? Or has the previous question been accepted as complete.
You can post a new one if you like. ;)
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@Deque Sure I would love too :-*
I don't know if this question has been posted earlier or not but if it is not then here you go.
Its a common and very easy question :
public class MonkeyDLuffy {
public static void main(String... string) {
System.out.println("Kaizoku-oni, orewa naru!");
//Character c = new Character('\u000d');
}
}
Does the above code compiles ? Give proper reasons.
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It can't compile, because at the beginning of the compiling/parsing progress every unicode sign will be read and replaced. In this case, \u000d, it will be replaced with a newline. This leads to a source code like:
public class MonkeyDLuffy {
public static void main(String... string) {
System.out.println("Kaizoku-oni, orewa naru!");
//Character c = new Character('
');
}
}
and some of the comment is now in a new line and not a comment anymore.
If my answer is correct someone else can ask the next question.
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It can't compile, because at the beginning of the compiling/parsing progress every unicode sign will be read and replaced. In this case, \u000d, it will be replaced with a newline. This leads to a source code like:
public class MonkeyDLuffy {
public static void main(String... string) {
System.out.println("Kaizoku-oni, orewa naru!");
//Character c = new Character('
');
}
}
and some of the comment is now in a new line and not a comment anymore.
If my answer is correct someone else can ask the next question.
Correct!
Make the next quiz.
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Hey guys,
I enjoy this kind of challenges and really would like to see them continue, so i thought it would be a nice way to introduce myself by posting one (since the last one was posted some time ago).
The challenge is probably too easy, simply explain what is happening ;)
import java.lang.reflect.Method;
public class Main {
private String key = "hello";
public Main(){
key = decrypt(key);
}
public static void main(String[] args) {
final Main m = new Main();
m.doStuff();
}
private String decrypt(final String s){
final char[] c = s.toCharArray();
for(int i = 0; i < c.length; i++){
c[i] = (char)(c[i] - 4 - i);
}
return String.valueOf(c);
}
private void hello(){
System.out.println("Hey guys!");
}
public void doStuff(){
try {
Method m = this.getClass().getDeclaredMethod(key);
m.invoke(this);
} catch (Exception e) {
//I don't care ;)
}
}
{{
key = "ljrsw";
}}
}