I'm little confused about how the generics works? I'm learning about function API in java and there I just test Function interface and got confused about compose method that how the generics is working in compose method.
Reading the generics on the java official tutorial website I realize that if we have any generic type in the method return or parameters we have to declare that type in the signature of method as explained below.
Here is the method I read in official docs tutorial.
public static <K, V> boolean compare(Pair<K, V> p1, Pair<K, V> p2) {
return p1.getKey().equals(p2.getKey()) &&
p1.getValue().equals(p2.getValue());
}
Above method have two types, K, V which are declared in the signature after the static keyword as but when I read java Function API there is one method called compose and the signature of the compose is as
default <V> Function<V, R> compose(Function<? super V, ? extends T> before) {
Objects.requireNonNull(before);
return (V v) -> apply(before.apply(v));
}
1) The first question where is the T & R declared? which are being used in the return type and in the parameter. Or my understanding is wrong?
Then I read more in generics tutorials and then I try to understand the concept of super and extends in generics and read here then I test compose method more and then confused again about how the super and extends works in the compose method?
public static void main(String... args){
Function<Integer, String> one = (i) -> i.toString();
Function<String, Integer> two = (i) -> Integer.parseInt(i);
one.compose(two);
}
As above I have declared two Function with lamdas. One is having Integer input and String output the other one is reversed from it.
2) The second question is that how Integer and String are related to extends and super? There is no relation between String and Integer class no one is extending each other then how it is working?
I tried my best to explain my question/problem. Let me know what you didn't understand I will try again.
Where are T and R defined?
Remember, compose is declared in the Function interface. It can not only use generic parameters of its own, but also the type's generic parameters. R and T are declared in the interface declaration:
interface Function<T, R> {
...
}
What are ? extends and ? super?
? is wildcard. It means that the generic parameter can be anything. extends and super give constraints to the wildcard. ? super V means that whatever ? is, it must be a superclass of V or V itself. ? extends T means that whatever ? is, it must be a subclass of T or T itself.
Now let's look at this:
Function<Integer, String> one = (i) -> i.toString();
Function<String, Integer> two = (i) -> Integer.parseInt(i);
one.compose(two);
From this, we can deduce that T is Integer and R is String. What is V? V must be some type such that the constraints Function<? super V, ? extends T> is satisfied.
We can do this by substituting the argument we passed in - Function<String, Integer> - to get String super V and Integer extends Integer.
The second constraint is satisfied already while the first constraint now says that String must be a super class of V or String itself. String cannot have subclasses so V must be String.
Hence, you can write something like:
Function<String, String> f = one.compose(two);
but not
Function<Integer, String> f = one.compose(two);
When you compose a Function<Integer, String> and a Function<String, Integer> you cannot possibly get a Function<Integer, String>. If you try to do this, V is automatically inferred to be Integer. But String super Integer is not satisfied, so the compilation fails. See the use of the constraints now? It is to avoid programmers writing things that don't make sense. Another use of the constraints is to allow you to do something like this:
Function<A, B> one = ...
Function<C, SubclassOfB> two = ...
Function<SubclassOfC, B> f = one.compose(two);
There is no relationship between Integer and String in this case, it's all about V.
1) The compose function is part of Interface Function<T,R>. As you can see in documentation for this interface:
Type Parameters:
T - the type of the input to the function
R - the type of the result of the function
2) The super and extends constraints in questions aren't applied to T & R, they're applied to the generic type parameters of a function that you pass in as an argument to the compose function.
Basically this means that if you have:
Function<ClassA, ClassB> one;
Function<SomeSuperClassOfC, SomeSubclassOfA> two;
then it's valid to call
Function<ClassC, ClassB> three = one.compose(two)
I will try to explain from zero;
interface Function<T, R> - this is interface with one method, which must be implemented R apply (T);
in Java prior to 8 we must write:
Function<Integer, String> one = new Function<Integer, String>() {
#Override
public String apply(Integer i) {
return i.toString();
}
};
now you can use it:
String resultApply = one.apply(5);
now, I think, you get the idea.
Related
Given the following class:
public class TestMap<K,V> {
private HashMap<K,V> map;
public void put(K k, V v) {
map.put(k, v);
}
public V get(K k) {
return map.get(k);
}
}
and this function:
static<T extends Object> String getObj(TestMap<T, String> m, String e) {
return m.get(e);
}
Why does "get" show this error:
The method get(T) in the type MyHashMap<T,String> is not applicable for the arguments (String)
When getObj states T extends Object, and the map has been initialized with TestMap<T, String> m, why can't String be used as parameter? I cant wrap my head around why this doesnt work, as String is a subtype of Object ?
I tried extending T from a custom superclass and using a subclass instead of String.
get() requires as an argument a key, which is of type T, not String.
So either you need to change e to type T or change the type of the Map, e.g., to TestMap<String, T>.
Generics describe what an Object is, List<String> strings is a list of Strings. That gives us some convenience, we don't have to cast. String s = strings.get(0); and we get some compile-time protection. Integer i = 1; then strings.add(i); fails to compile.
The error you are getting is a compile time protection. The key you're providing is an Object, so it might be the correct type. In the case of a map, "why not check?" It is just going to hashcode and check for equality which is ok. In the case of a List though, strings.add((Object)i); needs to fail.
If the doesn't fail, then later something might call. for(String s: strings) and get a class cast exception.
To fix this you can change you V get(K k) method to a V get(Object k) the same as map uses. Your get method requires type K a map normally has a get method with a Object
As mentioned "T extends Object" is the same as just "T". You use the extends when you want to put a lower bound on something. public <T extends Number> int sum(List<T> numbers). In this case "T" can be any class that extends numbers. We can do Number n = numbers.get(0); because we know it is at least a number. But we could not do numbers.add(n); because while T is at least a number, it could be something else like an Integer, Double, Number, BigDecimal etc.
The Function interface has the compose() and andThen() methods while the BiFunction interface only has the andThen() method. My question is simply how could the corresponding method be implemented? I'll try to represent this graphically.
The single letters are parameterized types as defined by Java's Function and BiFunction interfaces. Arrows are the flow of inputs and outputs. Boxes with connected arrows are functions. The dotted box just shows how the apply method is used.
The Function's Compose() and andThen() methods are straightforward since a Function has one input and one output and therefore can only be strung sequentially with another in two ways.
Since a BiFunction has one output, the "after" function has to be something with only one corresponding input, and Function fits the bill. And since it has two inputs, the "before" function needs to be something with two outputs? You can't have a method return two things, so there seemingly can't be a "before". The return type of each of these methods is the same as the interface they are defined in, so the proposed method should return a BiFunction.
My proposal then is a method that takes two Functions as input and returns a BiFunction. I'm not sure what else it could even be. It couldn't be two BiFunctions because then the return type would have to be a QuaterFunction.
Here is the code as it would be written in the Java Library:
public interface BiFunction<T, U, R> {
// apply()...
default <V, W> BiFunction<V, W, R> compose(
Function<? super V, ? extends T> beforeLeft,
Function<? super W, ? extends U> beforeRight) {
Objects.requireNonNull(beforeLeft);
Objects.requireNonNull(beforeRight);
return (V v, W w) -> apply(beforeLeft.apply(v), beforeRight.apply(w));
}
// andThen()...
}
Here is the finished graph:
Here it is in use:
BiFunction<Integer, Integer, Integer> add = Integer::sum;
Function<Integer, Integer> abs = Math::abs;
BiFunction<Integer, Integer, Integer> addAbs = add.compose(abs, abs);
System.out.println(addAbs.apply(-2, -3));
// output: 5
If you want to actually test this, you can do something like this:
public interface BiFunctionWithCompose<T, U, R> extends BiFunction<T, U, R> {...
Or like this:
package myutil;
public interface BiFunction<T, U, R> extends java.util.function.BiFunction<T, U, R> {...
I have no idea if this will be useful to anyone, but it was really fun to think through and write. Have a wonderful day.
How can I bind a Java Supplier to an existing instance of an Object? For example, if I want to write my own compareTo() method with this header:
public static int myCompareTo(Object o1, Object o2, Supplier<Comparable> supplier) {...}
I want be able to call it like:
myCompareTo("Hello", "Hello2", String::length);
where String (with the capital letter) is a class and no object. So how can I bind the instance o1 to the supplier?
Here's what you were searching for (I believe):
public static <T, U extends Comparable<U>> int compare(T o1, T o2, Function<T, U> mapper) {
return mapper.apply(o1).compareTo(mapper.apply(o2));
}
You can call that like so:
compare("str1", "str2", String::length); // 0
Thanks for your answers. Actually I figured it out now. I wanted to have the supplied object instances (o1 and o2) to execute the given method. I found out that Supplier was the wrong interface instead I had to use Function. Here you can see my working simplified example:
public static <T> int myCompareTo(T o1, T o2, Function<T, Comparable> getter) {
return getter.apply(o1).compareTo(getter.apply(o2));
}
The reason, the interface has to be Function and not Supplier is, that only Function is equivalent to a lambda expression taking an object and calls the referenced method on the object.
For example, if you define the method reference as:
Function<TypeOfInstance, ReturnTypeOfReferencedMethod> methodReference = TypeOfInstance::referencedMethod();
then the equivalent lambda expression being executed is:
(instance) -> instance.referencedMethod()
Additional Information:
Edit: I know I could have done the same by using Comparator, but this example is very simplified. In my application a Function of this kind is neccessary. I had to create a compareTo function that sorts an ArrayList by more than one attribute because the main sorting attribute may not be unique in the list. I want to share my code with you, because I think it can be a interesting insight for you.
public static <T> int ultimateCompare(T o1, T o2, Function<T, Comparable>... getters) {
for (Function<T, Comparable> getter : getters) {
int result = getter.apply(o1).compareTo(getter.apply(o2));
if (result != 0) return result;
}
return 0;
}
With this for example, you can sort a list of persons by last name and if two of them are identical, you can use the first name to sort. With this solution you can change sorting at runtime.
Actually a more correct way to define your method would be:
private static <T, U extends Comparable<? super U>> int myCompareTo(T left, T right, Function<T, U> fu) {
return Comparator.comparing(fu).compare(left, right);
}
You can use
Comparator.comparing(String::length);
to obtain a comparator instance which you can pass to the method.
While trying out a piece of code using generics, wildcard and functional programming, I have a doubt, A simple program below
import java.util.function.Function;
public static void main(String[] args) {
String x = "";
String y = getFormData().apply(x);
}
private static Function<?, String> getFormData() {
return x -> ((String) x).concat("asd");
}
The above program throws an error on the line
String y = getFormData().apply(x);
apply (capture<?>) in Function cannot be applied to (java.lang.String)
But if I change the getFormData() function to
private static Function<? super String, String> getFormData() {
return x -> ((String) x).concat("asd");
}
the error is gone.
Can somebody explain why this is happening ? I know <? super String> means, any super type of String or String itself.
The first version of your method:
Function<?, String> getFormData() {...}
could return a Function<Integer, String>. This function then clearly cannot be applied to a String!
In the second case however:
Function<? super String, String> getFormData() {...}
it is not allowed to return a Function<Integer, String>.
The type expression Function<?, String> is compatible with any Function that returns String, regardless of its argument type. For example, it is compatible with a Function<Integer, String>. Therefore, if you have no more specific information about it than that then you cannot safely pass a String -- or anything else -- as the argument to its apply method. The type expression does not provide enough information to determine what argument types are allowed.
On the other hand, the type expression Function<? super String, String> is compatible only with Functions that accept String or one of its supertypes as an argument (and return String). It is always safe to apply such a function to a String. It would also be safe to apply it to an instance of a subclass of String if it were possible to create such a subclass. It is not safe to apply it to any other type, however, not even a strict supertype of String, because the first type parameter of the Function could always be String itself.
Overall, since that means the only acceptable argument type is String, you don't appear to gain anything but confusion from using a wildcard. I'd just use Function<String, String> instead.
I'll try to illustrate my problem in the following simplified example:
public class DataHolder<T> {
private final T myValue;
public DataHolder(T value) {
myValue = value;
}
public T get() {
return myValue;
}
// Won't compile
public <R> DataHolder<R super T> firstNotNull(DataHolder<? extends R> other) {
return new DataHolder<R>(myValue != null ? myValue : other.myValue); }
public static <R> DataHolder<R> selectFirstNotNull(DataHolder<? extends R> first,
DataHolder<? extends R> second) {
return new DataHolder<R>(first.myValue != null ? first.myValue : second.myValue);
}
}
Here I want to write generic method firstNotNull that returns DataHolder parametrized by common supertype of type parameter T of the this and other argument, so later I could write e.g.
DataHolder<Number> r = new DataHolder<>(3).firstNotNull(new DataHolder<>(2.0));
or
DataHolder<Object> r = new DataHolder<>("foo").firstNotNull(new DataHolder<>(42));
The problem is that this definition of firstNotNull is rejected by compiler with message that super T part of type constraint is illegal (syntactically).
However without this constraint definition is also wrong (obviously), because in this case T and R are unrelated to each other.
Interestingly, definition of similar static method selectFirstNotNull is correct and the latter works as expected. Is it possible to achieve the same flexibility with non-static methods in Java type system at all?
It isn't possible to do this. The authors of Guava ran into the same issue with Optional.or. From that method's documentation:
Note about generics: The signature public T or(T defaultValue) is
overly restrictive. However, the ideal signature, public <S super T> S or(S), is not legal Java. As a result, some sensible operations
involving subtypes are compile errors:
Optional<Integer> optionalInt = getSomeOptionalInt();
Number value = optionalInt.or(0.5); // error
FluentIterable<? extends Number> numbers = getSomeNumbers();
Optional<? extends Number> first = numbers.first();
Number value = first.or(0.5); // error
As a workaround, it is always safe to cast an
Optional<? extends T> to Optional<T>. Casting either of the above
example Optional instances to Optional<Number> (where Number is the
desired output type) solves the problem:
Optional<Number> optionalInt = (Optional) getSomeOptionalInt();
Number value = optionalInt.or(0.5); // fine
FluentIterable<? extends Number> numbers = getSomeNumbers();
Optional<Number> first = (Optional) numbers.first();
Number value = first.or(0.5); // fine
Since DataHolder is immutable like Optional, the above workaround will work for you too.
See also: Rotsor's answer to Bounding generics with 'super' keyword
I don't think there is any easy and type-safe way to do this. I've tried a couple of approaches, but the only working approach that I found is to start with a super type generic instance, and make the method pretty simple like this:
public DataHolder<T> firstNotNull(DataHolder<? extends T> other) {
return new DataHolder<T>(myValue != null ? myValue : other.myValue);
}
Now you have to change your invocation to:
DataHolder<Number> r = new DataHolder<Number>(3).firstNotNull(new DataHolder<>(2.0));
You might argue that this doesn't really answer your question, but this is the simplest thing you're going to get, or better resort to a static method approach. You can surely come up with some highly convoluted (and type-unsafe) methods to do so, but readability should be of major concern here.
Try changing your method as follows:
public <R> DataHolder<R> firstNotNull(DataHolder<? super T> other) {
return new DataHolder<R>((R)(this.myValue != null ? myValue : other.myValue));
}
WARNING: This compiles and gives the appearance of being properly checked for the most part but is not perfect. It will restrict the input parameters, but not the output. This cannot be done perfectly. In some ways you might be better off doing this unchecked rather than giving the illusion of being checked. Here are some examples:
DataHolder<BigDecimal> a = new DataHolder<>(new BigDecimal(34.0));
DataHolder<Number> b = new DataHolder<>(new Integer(34));
DataHolder<String> c = new DataHolder<>("");
DataHolder<Number> p = a.firstNotNull(b); // WORKS (good)
DataHolder<BigDecimal> q = b.firstNotNull(a); // FAILS (good)
DataHolder<BigDecimal> r = b.firstNotNull(c); // FAILS (good)
DataHolder<String> s = a.firstNotNull(b); // WORKS (not good!!!)