Java Puzzlers NG S02: Down the Rabbit Hole as it was presented at The Pittsburgh Java Meetup Group

Developer Advocate @JFrog
@jbaruch on the internetz
Solution Architect @Confluent
@gAmUssA on the internetz

@tagir_valeev
@jbaruch http://jfrog.com/shownotes @gamussa

1. Two entertaining guys on
the stage
2. Funny puzzling questions
3. You think and vote
4.T-shirts are airborne
5. Official twitter hashtag:
#javapuzzlersng

Which Java version are you on?
A. Java 7
B. Java 8
C. Java 9
D. Java 6
E. Java 5
F. Java 2

Watching the puzzlers like… #dafaq

Everything works (or doesn't)
in the latest Java 8 and/or 9 update

public class PerfectRobbery {
private Semaphore bankAccount = new Semaphore(-42);
public static void main(String[] args) {
PerfectRobbery perfectRobbery = new PerfectRobbery();
perfectRobbery.takeAllMoney();
perfectRobbery.checkBalance();
}
public void takeAllMoney(){
bankAccount.drainPermits();
}
public void checkBalance(){
System.out.println(bankAccount.availablePermits());
}
}
A. IllegalArgumentException –can’t create semaphore with negative
B. UnsupportedOperationException –can’t drain when negative
C. 0
D. -42

A. IllegalArgumentException –can’t create semaphore with negative
B. UnsupportedOperationException –can’t drain when negative
C. 0
D. -42
public class PerfectRobbery {
private Semaphore bankAccount = new Semaphore(-42);
public static void main(String[] args) {
PerfectRobbery perfectRobbery = new PerfectRobbery();
perfectRobbery.takeAllMoney();
perfectRobbery.checkBalance();
}
public void takeAllMoney(){
bankAccount.drainPermits();
}
public void checkBalance(){
System.out.println(bankAccount.availablePermits());
}
}

Available
-42?!

A. true/true
B. true/false
C. false/true
D. false/false
Collections.emptyList() == Collections.emptyList();
Collections.emptyIterator() == Collections.emptyIterator();

A. true/true
B. true/false
C. false/true
D. false/false
Spliterators.emptySpliterator() == Spliterators.emptySpliterator();
Stream.empty() == Stream.empty();

Singleton Strikes Back!

A. true/true
B. true/false
C. false/true
D. false/false
Spliterators.emptySpliterator() == Spliterators.emptySpliterator();
Stream.empty() == Stream.empty();

Even empty Stream has state!

”Identical”
1. Has the same state
2. Not related to “equals and hashcode”contract
3. Not related to references to objects in memory

List[] twins = new List[2];
Arrays.setAll(twins, ArrayList::new);
A. Absolutely identical empty lists
B. Absolutely identical non-empty lists
C. Non-identical empty lists
D. Non-identical non-empty lists

How single is a Single Abstract Method Interface?

A. WTF?! ’Single’means one,not three!
B. Problem is with partyHard(T),remove it and it will work
C. Problem is the drinkIn methods,removing one of them and it will
work
D. It will work fine! Both partyHard() and drinkIn() are merged in
SingleAndHappy,leaving one abstract method
public interface Single<T> {
default void partyHard(String songName) { System.out.println(songName); }
void partyHard(T songName);
void drinkIn(T drinkName);
void drinkIn(String dringName);
}
@FunctionalInterface
public interface SingleAndHappy extends Single<String> { }

A. WTF?! ’Single’means one,not three!
B. Problem is with partyHard(T),remove it and it will work
C. Problem are the drinkIn methods,removing it will leave one abstract
method
D. Yes! Both partyHard() and drinkIn() are merged in SingleAndHappy,
leaving one abstract method
public interface Single<T> {
default void partyHard(String songName) { System.out.println(songName); }
void partyHard(T songName);
void drinkIn(T drinkName);
void drinkIn(String dringName);
}
public interface SingleAndHappy extends Single<String> { }

Hacking the bank
☑Bank software written in Java
☑Hack into it
☑Analyze the accounts

Given the code above,which statement is wrong:
A. The Set is ordered by hashcode
B. The order is predictable across multiple runs of the JVM on the same machine
C. The order of elements in Set is not predictable
D. Statements A & B are correct
Set<String> accounts= new HashSet<>(Arrays.asList("Gates", "Buffett", "Bezos", "Zuckerberg"));
System.out.println(”accounts= " + accounts);

A. The Set is ordered
Set<String> accounts= new HashSet<>(Arrays.asList("Gates", "Buffett", "Bezos", "Zuckerberg"));

public boolean add(E e) {
return map.put(e, PRESENT)==null;
}

Your turn,FBI

A. The Set is ordered
Set<String> accounts = Set.of("Gates", "Buffett", "Bezos", "Zuckerberg");

private int probe(Object pe) {
int idx = Math.floorMod(pe.hashCode() ^ SALT, elements.length);
while (true) {
E ee = elements[idx];
if (ee == null) {
return -idx - 1;
} else if (pe.equals(ee)) {
return idx;
} else if (++idx == elements.length) {
idx = 0;
}
}
}

What’s correct?
A. If you convert your application to module,classpath
dependencies will still be resolved correctly
B. If one of the dependencies was converted to a module,you
have to declare it in module-info in order to use
C. Once you added the module-info to your project you have to
declare the dependencies twice,in classpath and in module-
info
D. None of the above

A. You killed them all
B. You killed only even ones
C. They all survived
D. You killed only odd ones
E. All answers are correct
static void killThemAll(Collection<Hero> expendables) {
Iterator<Hero> heroes = expendables.iterator();
heroes.forEachRemaining(e -> {
if (heroes.hasNext()) {
heroes.next();
heroes.remove();
}
});
System.out.println(expendables);
}

Don’t do that.Really,don’t.
killThemAll(new ArrayList<String>(Arrays.asList("N","S","W","S","L","S","L","V")));
[]
killThemAll(new LinkedList<String>(Arrays.asList("N","S","W","S","L","S","L","V")));
[S,S,S,V]
killThemAll(new ArrayDeque<String>(Arrays.asList("N","S","W","S","L","S","L","V")));
[N,S,W,S,L,S,L,V]
killThemAll(new TreeSet<String>(Arrays.asList("N","S","W","S","L","S","L","V")));
[N,W,L,L]

Subtle Difference

A. Both work just fine
B. Lambda works,method ref fails
C. Method ref works,lambda fails
D. Won’t compile
public interface OriginalPredicate<T> {
boolean test(T t);
}
OriginalPredicate<Object> lambda = (Object obj) -> ”adidas".equals(obj);
OriginalPredicate<Object> methodRef = ”adidas"::equals;

A. Both work just fine
B. Lambda works,method ref fails
C. Method ref works,lambda fails
D. Not a functional interface,will fail on annotation processing
Public interface CopyCatPredicate {
<T> boolean test(T t);
}
CopyCatPredicate lambda = (Object obj) -> " adadas".equals(obj);
CopyCatPredicate methodRef = " adadas"::equals;

A generic function type for a functional interface may be implemented by a
method reference expression (§15.13), but not by a lambda expression
(§15.27) as there is no syntax for generic lambda expressions. “

A.[Data, Kirk, Spock]
B.[Spock, Kirk, Data, Data, Kirk, Spock]
C.[Spock, Kirk, Data]
D.[Data, Data, Kirk, Kirk, Spock, Spock]
E.Are you nuts? Won’t compile! Data with
Kirk?!
List<String> list = Stream.of("Spock", "Kirk", "Data", "Data", "Kirk",
"Spock").sequential()
.filter(new TreeSet<>()::add).collect(Collectors.toList());
System.out.println(list);

filter(new TreeSet<>()::add) filter(i -> new TreeSet<>().add(i))!=
New instance is
created every time!
Instance method is
created once!

A. obvious / obvious
B. obvious / NullPointerException
C. NullPointerException / obvious
D. NullPointerException / NullPointerException
Optional.of("obvious").orElseGet(null);
Optional.empty().map(null).orElse("obvious");

A. obvious / obvious
B. obvious / NullPointerException
C. NullPointerException / obvious
D. NullPointerException / NullPointerException
Optional.of("obvious").orElseGet(null);
Optional.empty().map(null).orElse("obvious");
Will never happen
Will never happen

Identical?

A. All
B. 3 and 4
C. Only 3
D. Other
1. Consumer<String> agentA = s -> System.out.println(s);
Consumer<String> agentB = s -> System.out.println(s);
2. Consumer<String> agentA = System.out::println;
Consumer<String> agentB = System.out::println;
3. Supplier<Consumer<String>> supplier = () -> s -> System.out.println(s);
Consumer<String> agentA = supplier.get();
Consumer<String> agentB = supplier.get();
4. Supplier<Consumer<String>> supplier = () -> System.out::println;
Consumer<String> agentA = supplier.get();
Consumer<String> agentB = supplier.get();
When agentA == agentB?

Reuse is only possible for pure functions
Consumers accept parameters == have
state
Supplier in 4 has state –the resolved
method reference

Conclusions

-Write readable code!
-Comment all the tricks
-Sometimes it’s just a bug
-Static code analysis FTW -
IntelliJ IDEA!
-RTFM!
-Don’t abuse lambdas and
streams!

-Trust us, we have much
more where those came
from.
-Puzzlers? Gotchas? Fetal
position inducing behavior?
-puzzlers@jfrog.com

-Shownotes!
-http://jfrog.com/shownotes
-Slides
-Video in 24 hours
-Links
-Ratings
-Raffle! (come early)

-Did you like it?
-Praise us on twitter!
-#javapuzzlersng
-@gamussa
-@jbaruch
-Didn’t like it?
-/dev/null

Java Puzzlers NG S02: Down the Rabbit Hole as it was presented at The Pittsburgh Java Meetup Group

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Java Puzzlers NG S02: Down the Rabbit Hole as it was presented at The Pittsburgh Java Meetup Group