有可能在Java中创建泛型类型的实例吗?我在想,根据我所看到的,答案是否定的(由于类型擦除),但如果有人能看到我遗漏的东西,我会很感兴趣:

class SomeContainer<E>
{
    E createContents()
    {
        return what???
    }
}

编辑:事实证明,超级类型令牌可以用来解决我的问题,但它需要大量基于反射的代码,如下面的一些答案所示。

我将把这个问题放一段时间,看看是否有人提出了与Ian Robertson的Artima文章截然不同的东西。


当前回答

如果你的意思是 新的E () 那就不可能了。我想补充一点,它并不总是正确的——你怎么知道E是否有公共的无参数构造函数? 但是你总是可以把创建委托给其他知道如何创建实例的类——它可以是class <E>或你的自定义代码,就像这样

interface Factory<E>{
    E create();
}    

class IntegerFactory implements Factory<Integer>{    
  private static int i = 0; 
  Integer create() {        
    return i++;    
  }
}

其他回答

正如您所提到的,您不能从泛型中获得实例。在我看来,你必须改变设计,使用FACTORY METHOD设计模式。通过这种方式,你不需要你的类或方法是泛型:

class abstract SomeContainer{
    Parent execute(){
        return method1();
    }
    
    abstract Parent method1();
}


class Child1 extends Parent{
    Parent method1(){
        return new Parent();
    } 
} 

class Child2 extends Parent{
    Parent method1(){
        return new Child2();
    } 
} 

正如你说的,你不能真正做到这一点,因为类型擦除。你可以使用反射来实现,但这需要大量的代码和错误处理。

package org.foo.com;

import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;

/**
 * Basically the same answer as noah's.
 */
public class Home<E>
{

    @SuppressWarnings ("unchecked")
    public Class<E> getTypeParameterClass()
    {
        Type type = getClass().getGenericSuperclass();
        ParameterizedType paramType = (ParameterizedType) type;
        return (Class<E>) paramType.getActualTypeArguments()[0];
    }

    private static class StringHome extends Home<String>
    {
    }

    private static class StringBuilderHome extends Home<StringBuilder>
    {
    }

    private static class StringBufferHome extends Home<StringBuffer>
    {
    }   

    /**
     * This prints "String", "StringBuilder" and "StringBuffer"
     */
    public static void main(String[] args) throws InstantiationException, IllegalAccessException
    {
        Object object0 = new StringHome().getTypeParameterClass().newInstance();
        Object object1 = new StringBuilderHome().getTypeParameterClass().newInstance();
        Object object2 = new StringBufferHome().getTypeParameterClass().newInstance();
        System.out.println(object0.getClass().getSimpleName());
        System.out.println(object1.getClass().getSimpleName());
        System.out.println(object2.getClass().getSimpleName());
    }

}

使用TypeToken<T>类:

public class MyClass<T> {
    public T doSomething() {
        return (T) new TypeToken<T>(){}.getRawType().newInstance();
    }
}

如果你在泛型类中需要一个类型参数的新实例,那么让你的构造函数要求它的类…

public final class Foo<T> {

    private Class<T> typeArgumentClass;

    public Foo(Class<T> typeArgumentClass) {

        this.typeArgumentClass = typeArgumentClass;
    }

    public void doSomethingThatRequiresNewT() throws Exception {

        T myNewT = typeArgumentClass.newInstance();
        ...
    }
}

用法:

Foo<Bar> barFoo = new Foo<Bar>(Bar.class);
Foo<Etc> etcFoo = new Foo<Etc>(Etc.class);

优点:

比Robertson的超级类型令牌(STT)方法简单得多(而且问题更少)。 比STT方法更有效(STT方法会把你的手机当早餐吃)。

缺点:

Can't pass Class to a default constructor (which is why Foo is final). If you really do need a default constructor you can always add a setter method but then you must remember to give her a call later. Robertson's objection... More Bars than a black sheep (although specifying the type argument class one more time won't exactly kill you). And contrary to Robertson's claims this does not violate the DRY principal anyway because the compiler will ensure type correctness. Not entirely Foo<L>proof. For starters... newInstance() will throw a wobbler if the type argument class does not have a default constructor. This does apply to all known solutions though anyway. Lacks the total encapsulation of the STT approach. Not a big deal though (considering the outrageous performance overhead of STT).