为什么使用泛型不会抛出运行时或编译时异常？

Question

I've got a method in a class that has a return type specified by use of a generic.

public class SomeMain {

  public static void main(String[] args) {

    Foo<Integer> foo = new Foo<Integer>();
    System.out.println(foo.getFoo()); // Works, prints out "Foo"

  }

  public static class Foo<E>  {
    public E getFoo() {
      return (E) "Foo";
    }
  }
}

With the generic return type, I assumed the return in the above example would evaluate to:

return (Integer) "Foo";  // Inconvertible types !!!

Instead a String is returned and printed correctly.

I get a compilation error if I change the call to be:

String fooString = foo.getFoo(); // Compile error, incompatible types found
System.out.println(fooString);

What am I missing to help me understand what's going on here and why the original version didn't result in a compilation error.

Answer 1

This is because overload resolution resolved your println call to println(Object) , since there is no println(Integer) .

Keep in mind that Java's generics are erased at runtime. And casts like (E) "Foo" are removed, and are moved to call site. Sometimes this is not necessary, so things are casted to the right type only when needed.

In other words, no casts are performed inside getFoo . The language spec supports this:

Section 5.5.2 Checked Casts and Unchecked Casts

• The cast is a completely unchecked cast.

  No run-time action is performed for such a cast.

After erasure, getFoo returns Object . And that gets passed into println(Object) , which is perfectly fine.

If I call this method and pass foo.getFoo , I will get an error:

static void f(Integer i) {
    System.out.println(i);
}
// ...
f(foo.getFoo()); // ClassCastException

because this time it needs to be casted.

Answer 2

System.out.println does not have an overload that takes Integer . So this statement:

System.out.println(foo.getFoo());

Is calling System.out.println(Object); .

To verify that it would otherwise fail, try:

Foo<Integer> foo = new Foo<Integer>();
Integer fooInt = foo.getFoo(); //class cast exception

The following will fail in the same way:

public static void main(String[] args) throws Exception {
    Foo<Integer> foo = new Foo<Integer>();
    print(foo.getFoo()); //Would also fail with a class cast exception
}
static void print(Integer in) {
    System.out.println(in);
}

And this is failing compilation for obvious reasons:

String fooString = foo.getFoo(); //can't work

foo is Foo<Integer> , and foo.getFoo() returns an Integer and the compiler can pick this up.

Answer 3

I'd like to add that during the type erasure process, the Java compiler replaces the unbounded type parameter E with Object , therefore the Foo class is actually compiled into:

public static class Foo {
    public Object getFoo() {
        return "Foo";
    }
}

That's why the following code is valid (cast is not needed):

Object obj = foo.getFoo();
System.out.println(obj);

At the same time, the next code snippet produces a compile-time error, as expected:

Foo<Integer> foo = new Foo<Integer>();
String fooString = foo.getFoo(); // you're trying to trick the compiler (unsuccessfully)
           ^
incompatible types: Integer can not be converted to String

And that's the main responsibility of generics - compile-time checks.

Yet there is another side of the story - execution-time casts. For example, if you write:

Integer value = foo.getFoo();

you get a ClassCastException thrown at runtime (the Java compiler inserts a checkcast instruction that examines whether the result of the foo.getFoo() can be cast to Integer ).