trampolines, monoids & other functional amenities This is NOT your father's by Mario Fusco mario.fusco@gmail.com twitter: @mariofusco Laziness,

public static void sort(List list, Comparator super T> c) Essence of Functional Programming Data and behaviors are the same thing! Data Behaviors Collections.sort(persons, (p1, p2) -> p1.getAge() – p2.getAge())

Higher-order functions Are they so mind-blowing?

Higher-order functions Are they so mind-blowing? … but one of the most influent sw engineering book is almost completely dedicated to them

Command Template Method Functions are more general and higher level abstractions Factory Strategy

public interface Converter { double convert(double value); } public class AbstractConverter implements Converter { public double convert(double value) { return value * getConversionRate(); } public abstract double getConversionRate(); } public class Mi2KmConverter extends AbstractConverter { public double getConversionRate() { return 1.609; } } public class Ou2GrConverter extends AbstractConverter { public double getConversionRate() { return 28.345; } } A strategy pattern Converter

public List convertValues(List values, Converter converter) { List convertedValues = new ArrayList(); for (double value : values) { convertedValues.add(converter.convert(value)); } return convertedValues; } List values = Arrays.asList(10, 20, 50); List convertedDistances = convertValues(values, new Mi2KmConverter()); List convertedWeights = convertValues(values, new Ou2GrConverter()); Using the Converter

A functional Converter public class Converter implements ExtendedBiFunction { @Override public Double apply(Double conversionRate, Double value) { return conversionRate * value; } } @FunctionalInterface public interface ExtendedBiFunction extends BiFunction { default Function curry1(T t) { return u -> apply(t, u); } default Function curry2(U u) { return t -> apply(t, u); } }

Currying Converter converter = new Converter(); double tenMilesInKm = converter.apply(1.609, 10.0); Function mi2kmConverter = converter.curry1(1.609); double tenMilesInKm = mi2kmConverter.apply(10.0); Converter value rate result Mi2km Converter value rate=1.609 result curry1 List values = Stream.of(10, 20, 50) .map(new Converter().curry1(1.609)) .collect(toList())

Function Composition Celsius  Fahrenheit : F = C * 9/5 + 32 Converter value rate=9/5

Function Composition Celsius  Fahrenheit : F = C * 9/5 + 32 Converter value rate=9/5 andThen n -> n+32 result

Function Composition Celsius  Fahrenheit : F = C * 9/5 + 32 Converter value rate=9/5 andThen n -> n+32 result Celsius2FarenheitConverter Function c2fConverter = new Converter().curry1(9.0/5) .andThen(n -> n + 32);

More Function Composition @FunctionalInterface public interface ExtendedBiFunction extends BiFunction { default ExtendedBiFunction compose1(Function super V, ? extends T> before) { return (v, u) -> apply(before.apply(v), u); } default ExtendedBiFunction compose2(Function super V, ? extends U> before) { return (t, v) -> apply(t, before.apply(v)); } } default Function compose(Function super V, ? extends T> before) { return (V v) -> apply(before.apply(v)); }

More Function Composition Fahrenheit  Celsius : C = (F - 32) * 5/9 Converter rate=5/9 result

More Function Composition Fahrenheit  Celsius : C = (F - 32) * 5/9 Converter rate=5/9 value n -> n-32 result compose2

More Function Composition Fahrenheit  Celsius : C = (F - 32) * 5/9 Converter rate=5/9 value n -> n-32 result Farenheit2CelsiusConverter Function f2cConverter = new Converter().compose2((Double n) -> n - 32) .curry1(5.0/9); Functions are building blocks to create other functions compose2

Monoids A monoid is a triple (T, , ∗ z) such that ∗ is an associative binary operation on T, and z ∈ T has the property that for all x ∈ T it holds that x∗z = z∗x = x. interface Monoid { T append(T a, T b); T zero(); } class Appender implements Monoid { public String append(String a, String b) { return a + b; } public String zero() { return ""; } } class Multiplier implements Monoid { public Integer append(Integer a, Integer b) { return a * b; } public Integer zero() { return 1; } }

Endomorphisms & Monoids interface Endomorphism extends Function { } interface EndoMonoid extends Monoid> { @Override default Endomorphism append(Endomorphism f1, Endomorphism f2) { return ??? } @Override default Endomorphism zero() { return ??? } }

Endomorphisms & Monoids interface Endomorphism extends Function { } interface EndoMonoid extends Monoid> { @Override default Endomorphism append(Endomorphism f1, Endomorphism f2) { return ??? } @Override default Endomorphism zero() { return ??? } } f1.andThen(f2); Function.identity();

public class SalaryCalculator { // B = basic + 20% public double plusAllowance(double d) { return d * 1.2; } // C = B + 10% public double plusBonus(double d) { return d * 1.1; } // D = C - 30% public double plusTax(double d) { return d * 0.7; } // E = D - 10% public double plusSurcharge(double d) { return d * 0.9; } public double calculate(double basic, boolean[] flags) { double salary = basic; if (flags[0]) salary = plusAllowance(salary); if (flags[1]) salary = plusBonus(salary); if (flags[2]) salary = plusTax(salary); if (flags[3]) salary = plusSurcharge(salary); return salary; } } SalaryCalculator

public class FluentEndoMonoid implements EndoMonoid { private final Endomorphism endo; public FluentEndoMonoid(Endomorphism endo) { this.endo = endo; } public FluentEndoMonoid(Endomorphism endo, boolean b) { this.endo = b ? endo : zero(); } public FluentEndoMonoid add(Endomorphism other) { return new FluentEndoMonoid(append(endo, other)); } public FluentEndoMonoid add(Endomorphism other, boolean b) { return add(b ? other : zero()); } public Endomorphism get() { return endo; } public static FluentEndoMonoid endo(Endomorphism f, boolean b) { return new FluentEndoMonoid(f, b); } } FluentEndoMonoid

public class SalaryCalculator { public double calculate(double basic, boolean [] flags) { return getCalculator(bs).apply(basic); } public Endomorphism getCalculator(boolean[] flags) { return endo(this::plusAllowance, flags[0]) .add(this::plusBonus, flags[1]) .add(this::plusTax, flags[2]) .add(this::plusSurcharge, flags[3]) .get(); } } Endomorphism f = salaryCalc.getCalculator(true, false, false, true); double aliceNet = f.apply(alice.getIncome()); double brianNet = f.apply(brian.getIncome()); Functional SalaryCalculator You can calculate a single salary … … but also obtain a calculator for a given combination of flags (Factory)

Lazy Evaluation Lazy evaluation (or call-by-name) is an evaluation strategy which delays the evaluation of an expression until its value is needed I know what to do. Wake me up when you really need it

Creating a Stream of prime numbers public static IntStream primes(int n) { return IntStream.iterate(2, i -> i + 1) .filter(n –> isPrime(n)) .limit(n); } public static boolean isPrime(int candidate) { int candidateRoot = (int) Math.sqrt((double) candidate); return IntStream.rangeClosed(2, candidateRoot) .noneMatch(i -> candidate % i == 0); }

Creating a Stream of prime numbers public static IntStream primes(int n) { return IntStream.iterate(2, i -> i + 1) .filter(n –> isPrime(n)) .limit(n); } public static boolean isPrime(int candidate) { int candidateRoot = (int) Math.sqrt((double) candidate); return IntStream.rangeClosed(2, candidateRoot) .noneMatch(i -> candidate % i == 0); } It iterates through every number every time to see if it can be exactly divided by a candidate number, but it would be enough to only test numbers that have been already classified as prime Inefficient

Recursively creating a Stream of primes static Intstream numbers() { return IntStream.iterate(2, n -> n + 1); } static int head(IntStream numbers) { return numbers.findFirst().getAsInt(); } static IntStream tail(IntStream numbers) { return numbers.skip(1); } static IntStream primes(IntStream numbers) { int head = head(numbers); return IntStream.concat( IntStream.of(head), primes(tail(numbers).filter(n -> n % head != 0)) ); }

Recursively creating a Stream of primes static Intstream numbers() { return IntStream.iterate(2, n -> n + 1); } static int head(IntStream numbers) { return numbers.findFirst().getAsInt(); } static IntStream tail(IntStream numbers) { return numbers.skip(1); } static IntStream primes(IntStream numbers) { int head = head(numbers); return IntStream.concat( IntStream.of(head), primes(tail(numbers).filter(n -> n % head != 0)) ); } Cannot invoke 2 terminal operations on the same Streams Problems? No lazy evaluation in Java leads to an endless recursion

Lazy evaluation in Scala def numbers(n: Int): Stream[Int] = n #:: numbers(n+1) def primes(numbers: Stream[Int]): Stream[Int] = numbers.head #:: primes(numbers.tail filter (n -> n % numbers.head != 0)) lazy concatenation In Scala the #:: method (lazy concatenation) returns immediately and the elements are evaluated only when needed

interface HeadTailList { T head(); LazyList tail(); default boolean isEmpty() { return true; } } Implementing a lazy list in Java class LazyList implements HeadTailList { private final T head; private final Supplier> tail; public LazyList(T head, Supplier> tail) { this.head = head; this.tail = tail; } public T head() { return head; } public HeadTailList tail() { return tail.get(); } public boolean isEmpty() { return false; } }

… and its lazy filter public HeadTailList filter(Predicate p) { return isEmpty() ? this : p.test(head()) ? new LazyList<>(head(), () -> tail().filter(p)) : tail().filter(p); } 2 3 4 5 6 7 8 9 2 3 5 7

Back to generating primes static HeadTailList primes(HeadTailList numbers) { return new LazyList<>( numbers.head(), () -> primes(numbers.tail() .filter(n -> n % numbers.head() != 0))); } static LazyList from(int n) { return new LazyList(n, () -> from(n+1)); } LazyList numbers = from(2); int two = primes(numbers).head(); int three = primes(numbers).tail().head(); int five = primes(numbers).tail().tail().head();

Printing primes static void printAll(HeadTailList list) { while (!list.isEmpty()){ System.out.println(list.head()); list = list.tail(); } } printAll(primes(from(2))); static void printAll(HeadTailList list) { if (list.isEmpty()) return; System.out.println(list.head()); printAll(list.tail()); } printAll(primes(from(2))); iteratively recursively

Iteration vs. Recursion  External Iteration public int sumAll(int n) { int result = 0; for (int i = 0; i <= n; i++) { result += i; } return result; }  Recursion public int sumAll(int n) { return n == 0 ? 0 : n + sumAll(n - 1); }  Internal Iteration public static int sumAll(int n) { return IntStream.rangeClosed(0, n).sum(); }

public class PalindromePredicate implements Predicate { @Override public boolean test(String s) { return isPalindrome(s, 0, s.length()-1); } private boolean isPalindrome(String s, int start, int end) { while (start < end && !isLetter(s.charAt(start))) start++; while (start < end && !isLetter(s.charAt(end))) end--; if (start >= end) return true; if (toLowerCase(s.charAt(start)) != toLowerCase(s.charAt(end))) return false; return isPalindrome(s, start+1, end-1); } } Another Recursive Example Tail Rescursive Call

What's the problem? List sentences = asList( "Dammit, I’m mad!", "Rise to vote, sir!", "Never odd or even", "Never odd and even", "Was it a car or a cat I saw?", "Was it a car or a dog I saw?", VERY_LONG_PALINDROME ); sentences.stream() .filter(new PalindromePredicate()) .forEach(System.out::println); Exception in thread "main" java.lang.StackOverflowError at java.lang.Character.getType(Character.java:6924) at java.lang.Character.isLetter(Character.java:5798) at java.lang.Character.isLetter(Character.java:5761) at org.javaz.trampoline.PalindromePredicate.isPalindrome(PalindromePredicate.java:17) at org.javaz.trampoline.PalindromePredicate.isPalindrome(PalindromePredicate.java:21) at org.javaz.trampoline.PalindromePredicate.isPalindrome(PalindromePredicate.java:21) at org.javaz.trampoline.PalindromePredicate.isPalindrome(PalindromePredicate.java:21) ……..

Tail Call Optimization int func_a(int data) { data = do_this(data); return do_that(data); } ... | executing inside func_a() push EIP | push current instruction pointer on stack push data | push variable 'data' on the stack jmp do_this | call do_this() by jumping to its address ... | executing inside do_this() push EIP | push current instruction pointer on stack push data | push variable 'data' on the stack jmp do_that | call do_that() by jumping to its address ... | executing inside do_that() pop data | prepare to return value of 'data' pop EIP | return to do_this() pop data | prepare to return value of 'data' pop EIP | return to func_a() pop data | prepare to return value of 'data' pop EIP | return to func_a() caller ...

Tail Call Optimization int func_a(int data) { data = do_this(data); return do_that(data); } ... | executing inside func_a() push EIP | push current instruction pointer on stack push data | push variable 'data' on the stack jmp do_this | call do_this() by jumping to its address ... | executing inside do_this() push EIP | push current instruction pointer on stack push data | push variable 'data' on the stack jmp do_that | call do_that() by jumping to its address ... | executing inside do_that() pop data | prepare to return value of 'data' pop EIP | return to do_this() pop data | prepare to return value of 'data' pop EIP | return to func_a() pop data | prepare to return value of 'data' pop EIP | return to func_a() caller ... caller avoid putting instruction on stack

from Recursion to Tail Recursion  Recursion public int sumAll(int n) { return n == 0 ? 0 : n + sumAll(n - 1); }  Tail Recursion public int sumAll(int n) { return sumAll(n, 0); } private int sumAll(int n, int acc) { return n == 0 ? acc : sumAll(n – 1, acc + n); }

Tail Recursion in Scala def isPalindrome(s: String): Boolean = isPalindrome(s, 0, s.length-1) @tailrec def isPalindrome(s: String, start: Int, end: Int): Boolean = { val pos1 = nextLetter(s, start, end) val pos2 = prevLetter(s, start, end) if (pos1 >= pos2) return true if (toLowerCase(s.charAt(pos1)) != toLowerCase(s.charAt(pos2))) return false isPalindrome(s, pos1+1, pos2-1) } def nextLetter(s: String, start: Int, end: Int): Int = if (start > end || isLetter(s.charAt(start))) start else nextLetter(s, start+1, end) def prevLetter(s: String, start: Int, end: Int): Int = if (start > end || isLetter(s.charAt(end))) end else prevLetter(s, start, end-1)

Tail Recursion in Java?  Scala (and many other functional languages) automatically perform tail call optimization at compile time  @tailrec annotation ensures the compiler will optimize a tail recursive function (i.e. you will get a compilation failure if you use it on a function that is not really tail recursive)  Java compiler doesn't perform any tail call optimization (and very likely won't do it in a near future) How can we overcome this limitation and have StackOverflowError-free functions also in Java tail recursive methods?

Trampolines to the rescue A trampoline is an iteration applying a list of functions. Each function returns the next function for the loop to run. Func1 return apply Func2 return apply Func3 return apply FuncN apply … result return

Implementing the TailCall … @FunctionalInterface public interface TailCall { TailCall apply(); default boolean isComplete() { return false; } default T result() { throw new UnsupportedOperationException(); } default T invoke() { return Stream.iterate(this, TailCall::apply) .filter(TailCall::isComplete) .findFirst() .get() .result(); } // ... missing terminal TailCall }

… and the terminal TailCall public static TailCall done(final T value) { return new TailCall() { @Override public boolean isComplete() { return true; } @Override public T result() { return value; } @Override public TailCall apply() { throw new UnsupportedOperationException(); } }; }

Using the Trampoline public class PalindromePredicate implements Predicate { @Override public boolean test(String s) { return isPalindrome(s, 0, s.length()-1).invoke(); } private TailCall isPalindrome(String s, int start, int end) { while (start < end && !isLetter(s.charAt(start))) start++; while (end > start && !isLetter(s.charAt(end))) end--; if (start >= end) return done(true); if (toLowerCase(s.charAt(start)) != toLowerCase(s.charAt(end))) return done(false); int newStart = start + 1; int newEnd = end - 1; return () -> isPalindrome(s, newStart, newEnd); } }

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Mario Fusco Red Hat – Senior Software Engineer mario.fusco@gmail.com twitter: @mariofusco Q A Thanks … Questions?