为什么在返回类型之前为静态泛型方法键入参数

Question

The following noGood method gives a compilation error because it omits the formal type parameter immediately before the return type T .

public static T noGood(T t) {
  return t;
}

Could somebody please help me understand that why is it required for a static generic method to have a type parameter before the return type? Is it not required for a non-static method?

Answer 1

The type parameter ( T ) is declared when you instantiate the class. Thus, instance methods don't need a type argument, since it's defined by the instance.

static methods, on the other hand, don't belong to an instance - they belong to the class. Since there's no instance to get the type information from, it must be specified for the method itself.

Answer 2

T wasn't defined. The order of modifiers and the return type remains the same.

public static <T> T noGood(T t) {
    return t;
}

Answer 3

When you're using generic, you need to declare them, using <> notation

1. In a class

    public class Foo<T, U, V>{ } 

2. In a method, before the return type

    public static <T, U extends Number, V> T foo(T t) { U u = ..; ... } public static <T> int foo(T t) { ... } 

Answer 4

First of all, this is pretty standard in languages. Even in C++:

template <class myType>
myType GetMax (myType a, myType b) {
    return (a>b?a:b);
}

You declare the type parameter above a generic function.

When its a member function of a class, it has access to the class's type parameters. When its static it doesn't, so you need to declare them explicitly.