如何在 Java 中推断出泛型类型？

Question

The Function.identity() returns a function, Function<T, T> that always returns its input argument(ie identity function ).

But being a static method, how does it know which concrete argument to return in place of type parameter T when it doesn't even take any inputs?

Illustration of my thought process:

Map idToPerson = people.collect( Collectors.toMap( (person -> person.getID() , Function.identity() ) );

Question: So how does the compiler figure out that Function.identity() is supposed to return Function<element of 'people' stream, element of 'people' stream> stream despite having no inputs?

---

According to the OpenJDK , the implementation is something like:

static <T> Function<T, T> identity()                                                                                                                          
{ 
    return t -> t;    
}

An attempt to narrow down my question:
How does Function.identity() know what the concrete data type t in t -> t (btw this is the lambda Function<T, T> ) is?

Answer 1

The Java type inference algorithm is based on the resolution of constraint formulas on inference variables . It is described in detail in Chapter 18 of the Java Language Specification . It's a bit involved.

Informally, for the example above the reasoning would go roughly as follows:

We have an invocation of Function.<T>identity() . Because most type parameters are named T , and consistently with the JLS, I'll use Greek letters to denote inference variables. So in this initial expression T:: α . What constraints do we have on α ?

Well identity() returns an instance of Function<α,α> used as argument to toMap .

static <T,K,U> Collector<T,?,Map<K,U>>  toMap(Function<? super T,? extends K> keyMapper, 
   Function<? super T,? extends U> valueMapper)

So now we have the constraints {α:> T, α <: K} (where :> means supertype of and vice-versa). This now requires us to infer T and K in this expression, which we'll refer to as β and γ , so: {α:> β, α <: γ} . To avoid getting bogged down in details, let's work through β only.

toMap then returns a collector as argument to Stream.collect , which provides us with another source of constraints:

collect(Collector<? super T,A,R> collector)

So now we know that {β:> T} . But here T also needs to be inferred, so it becomes an inference variable, and we have {β:> δ} .

This is where it starts unfolding, because the type parameter T for method collect refers to the parameter T in Stream<T> . So assuming the stream was defined as Stream<Person> , now we have {δ=Person} and we can reduce as follows:

• {β:> δ} => {β:> Person} ( β is a supertype of Person );
• {α:> β} => {α:> (β:> Person)} => {α:> Person)} ( α is a supertype of Person );

So through the process of inference we figured out that the type variable for Function.identity needs to be Person or a supertype of Person . A similar process for α <: γ would yield {α <: Person} (if the return type is specified). So we have two constraints:

• α needs to be Person or a supertype of Person ;
• α needs to be Person or a subtype of Person ;

Clearly the only type that satisfies all these constraints is Person .