Question

有一种方法可以避免慢速反射从一个类创建一个实例，显然在另一个方法中？例如：

Foo foo = new Foo();
foo.create(Dog.class, "rocky");

class Foo {
    Object create(Class object, String dogName) {
        //create an instance of the class 'object' here passing the argument to constructor
        //e.g. Object obj = new object(dogName); <-- this is wrong

        return obj;
    }
}

class Dog extends Animal {
   Dog(String dogName) {
       this.name = dogName;
   }
}

class Animal {
    String name;
}

我无法使用关键字＆＃34; new＆＃34;创建实例，因为我必须以动态方式在另一个方法中创建实例...

你有全权改善这个代码的最佳方式（如性能）:)谢谢！

Answer 1

让您的狗和动物类保持不变。并使用Builder模式

  public interface Builder<T> {
  public T build(String nameString);
}
public static void main(String[] args){
  Builder<Dog> builder =  new Builder<Dog>()
  {

     @Override
     public Dog build(String nameString)
     {
        return new Dog(nameString);
     }

  };
 Dog dog = builder.build("Rocky");
 System.out.print(dog.name);

}

Instantiating object of type parameter的答案也会进一步解释。

Answer 2

public <T> T create(Class<T> myClass, String constructorArg) 
throws IllegalArgumentException, InstantiationException, IllegalAccessException, InvocationTargetException, SecurityException, NoSuchMethodException {
    Constructor<T> toCall = myClass.getConstructor(String.class);
    return toCall.newInstance(constructorArg);
}

如何以最佳方式创建类对象中的实例

2 个答案: