Functional Programming with Java 8 Functions

Ok, let's start with something simple. The following is a lambda expression (i.e. an anonymous function) that takes an argument x and increments it by one. In other words a function that receives, apparently an integer, and returns a new incremented integer:

x -> x + 1

And what is the type of this function in Java?

Well, the answer is that it depends. In Java the same lambda expression could be bound to variables of different types. For instance, the following two are valid declarations in Java:

Function<Integer,Integer> add1 = x -> x + 1;
Function<String,String> concat = x -> x + 1;

The first one increments an integer x by one, whereas the second one concatenates the integer 1 to any string x.

And how can we invoke these functions?

Well, now that they are bound to a reference we can treat them pretty much like we treat any object:

Integer two = add1.apply(1); //yields 2
String answer = concat1.apply("0 + 1 = "); //yields "0 + 1 = 1"

So, as you can see every function has a method apply that we use to invoke it and pass it an argument.

And what if I already have a method that does that, can I use it as a function?

Yes, since an alternative way to create functions is by using methods we had already defined and that are compatible with our function definition.

Suppose that we have the following class definition with methods as defined below:

public class Utils {
   publicstatic Integer add1(Integer x) { return x + 1; }
   publicstatic String concat1(String x) { return x + 1; }
}

As you can see the methods in this class are compatible with our original function definitions, so we could use them as "method references" to create the same functions we did before with lambda expressions.

Function<Integer,Integer> add1 = Utils::add1;
Function<String,String> concat1 = Utils::concat1;

These two are just the same thing as the ones we did before.

High Order Programming

The cool thing about functions is that they encapsulate behavior, and now we can take a piece of code, put it into a function and pass it around to other functions or methods for them to use it. This type of functions that operate on (or produce new) functions are typically called high order functions and the programming style based on exploiting this powerful feature is called, unsurprisingly, high order programming.

About Functions that Create Functions

Let's see a couple of examples of how we can do this using function objects. Let's consider the following example:

Function<Integer, Function<Integer,Integer>> makeAdder = x -> y -> x + y;

Above we have a function called makeAdder that takes an integer x and creates a new function that takes an integer y and when the latter is invoked, it adds xto y. We can tell this is a high order function because it produces a new function.

Now, we use this to create a new version of the add1 function:

Function<Integer,Integer> add1 = makeAdder.apply(1);
Function<Integer,Integer> add2 = makeAdder.apply(2);
Function<Integer,Integer> add3 = makeAdder.apply(3);

With our high order function, however, we could also create add2, add3, ..., addn, right?

Can't we define this in a simpler way as we did before with the Utils class methods?

Yes, we can. Consider the following addition to the Utils class:

public class Utils {
    public static Function<Intger, Integer> adder(Integer x) {
       return y -> x + y;
    }
}

This signature is a bit simpler to read than that in our lambda expression, but as you can see it's pretty much the same thing. The function continues to receive the same number and type of arguments and continues to return the same type of result.

We can now use this simpler function factory to create our makeAdded and add1functions again:

Function<Integer, Function<Integer,Integer>> makeAdder = Utils::adder;
Function<Integer,Integer> add1 = makeAdder.apply(1);

And there we have it again, this is exactly the same thing as we had before.

About Functions that Receive Functions as Arguments

Let's suppose we had the following two functions defined:

Function<Integer,Integer> add1 = x -> x + 1;
Function<Integer,Integer> mul3 = x -> x * 3;

Now, naively, we could invoke this two functions together to increment and multiply a number by 3, right?. Like this:

Integer x = 10;
Integer res = mul3.apply(add1.apply(x)); //yields 33

But what if we created a function that did both things instead?

Consider the following pseudocode:

(f,g) -> x -> g( f(x) )

This would be a function that takes two other unary functions and creates yet another function that applies the original two in certain order. This is a classical example of what is called function composition.

In some languages there is even a binary operator to compose two functions in this way:

h = f o g

Where o would be an operator that would compose functions f and g pretty much as we did in pseudocode above and produce a new function h.

How can we do function composition in Java?

I can think of two ways to do this in Java, one more difficult than the other. Let's start with the more difficult strategy first, because that will let us appreciate the value of the simpler solution later on.

Function Composition Strategy 1

First, we must start by realizing that the function in pseudocode above is a binary function (i.e. a function that receives two arguments). But all our examples so far have dealt only with unary functions.

It would seem this is not important, but in Java it is, since functions of different arities have different target functional interfaces. In Java, a function that receives two arguments is called BinaryOperator.

For instance, using a BinaryOperator we could implement a sum operator:

BinaryOperator<Integer> sum = (a,b) -> a + b;
Integer res = sum.apply(1,2); // yields 3

Well, just as easily we could implement the compose operator, right? Only that in this case, instead of two simple integers, we receive two unary functions:

BinaryOperator<Function<Integer,Integer>> compose = (f,g) -> x -> g.apply(f.apply(x));

Now we can easily use this to fabricate a compund function that adds 1 and multiplies by 3, one after the other.

Function<Integer,Integer> h = compose.apply(add1,mul3);
Integer res = h.apply(10); //yields 33

And now we can beautifully, and in really simple way, combine two unary integer functions.

Function Composition Strategy 2

Now, function composition is something so common that it would have been a mistake if the Java Expert Group would have not considered it in their API design, and so, to make our lives simpler, all Function objects have a method called compose that allows us to very easily compose two functions together.

The following code produces the exact same result as above:

Function<Integer,Integer> h = mul3.compose(add1);
Integer res = h.apply(10);

Partial Function Application or Currying

In most functional programming languages it is possible to create partially applied functions. That is, if a function is receiving multiple arguments, we can partially invoke the function providing just a few arguments and receive a partially applied function out of it. This is typically called currying.

Although you have not noticed it, we have already covered that in this article, but now we are going to make it much more evident :-)

So, consider the following pseudocode

sum = x -> y -> x + y

Then we say that sum is a function that accepts one parameter x and fabricates another anonymous function that, in turn, accepts one parameter y that, when invoked, sums x and y.

In many functional programming languages a construct like this can be invoked as if this was just one simple function by doing:

sum 10 5 //yields 15

But the truth is that this is just syntactic sugar to do:

sum(10)(5) //yields 15

Since sum is a function that returns a function.

The beauty of this idiom is that now we could partially apply sum:

plus10 = sum 10

And now plus10 is a partially applied function of sum, bound to a first argument 10, and yet expecting a second argument y.

Can you see now where we had already talked about a similar idea in this article?

plus10(5) //yields 15

Can we do this with Java?

The truth is that we have already done it above, we just probably did not notice. Unfortunately, in Java we do not have the syntactic sugar that some other language have, and therefore this is a bit more verbose:

Function<Integer, Function<Integer, Integer>> sum = x -> y -> x + y;

Well, you can see sum is declared in a "currified" way. And now we can partially apply it:

Function<Integer, Integer> plus10 = sum.apply(10);
Integer res = plus10.apply(5); //yields 15

Unary Functions

So, as mentioned above, Java uses different functional interfaces for different function arities. And so Function<T,R> is a functional interface for any unary function where the domain and the codomain of the function may be of different types.

For instance, we could define a function that receives a string value and parses it as an integer:

Function<String,Integer> atoi = s -> Integer.valueOf(s);

But most of our examples above are for integer functions whose argument and return value are of this same type. For those cases we could alternatively use the UnaryOperator<T> instead. This is just a Function<T,T>.

Thus, some our declarations above could be slightly simplified with this functional interface:

UnaryOperator<Integer> add1 = n -> n + 1;
UnaryOperator<String> concat1 = s -> s + 1;
Function<Integer, UnaryOperator<Integer>> sum = x -> y -> x + y;
UnaryOperator<Integer> sum10 = sum.apply(10);

I have already written another article that explains Why Java 8 has Interface Pollution like this in case you are interested in an explanation.

Value Types and Primitive Type Functions?

Evidently using a type like Integer incurs into the costs of boxing and unboxing when our functions have to deal with a primitive type like int.

As you know, Java does not support value types as type arguments in generic declarations, so to deal with this problem we can use alternative functional interfaces like ToIntFunction, IntFunction or IntUnaryOperator.

Or we can define our own primitive function.

interface IntFx {
    publicintapply(intvalue);
}

Then we can do:

IntFx add1 = n -> n + 1;

IntFunction<IntFx> sum = x -> y -> x + y;
IntFx sum10 = sum.apply(10);
sum10.apply(4); //yields 14

Similar functional interfaces can be found for types double and long as well. This topic is also covered in the alternative article mentioned above about interface pollution.

Further Reader

• High Order Programming
• High Order Functions
• Function Composition
• Currying
• Java 8 Interface Pollution