泛型简单总结

作者: 机智的黑猫 | 来源:发表于2020-02-27 18:23 被阅读0次

泛型简单总结
Java泛型
泛型、型变与投影
泛型简单的应用总结：
泛型
JDK1.5后增加了泛型，那么为什么要有泛型呢？我们该如何自定义
java基础-泛型
Java泛型总结
Java笔记---泛型
泛型总结

为了方便理解首先先创建3个类

public class A {
        public void printA() {

        }
    }

    public class B extends A {
        public void printB() {

        }
    }

    public class C extends B {
        public void printC() {

        }
    }

 public class GenericHolder<T> {
        T t;

        public void putItem(T item) {
            t = item;
        }

        public T getItem() {
            return t;
        }

    }

如果希望只取出，不插入，或者是方法的返回值是泛型参数没有就使用? extends B
也叫做上边界通配符，协变

       GenericHolder<? extends B> holder1 = new GenericHolder<>();
        A a = new A();
        B b = new B();
        C c = new C();

        holder1.putItem(a);//编译错误
        holder1.putItem(b);//编译错误
        holder1.putItem(c);//编译错误


        A a1 = holder1.getItem();
        B b1 = holder1.getItem();
        C c1 = holder1.getItem();//编译错误

        val holder1:GenericHolder<out B> = GenericHolder()
        val a = A()
        val b = B()
        val c = C()
        holder1.putItem(a)//编译错误
        holder1.putItem(b)//编译错误
        holder1.putItem(c)//编译错误

        val a1: A? = holder1.item
        val b1: B? = holder1.item
        val c1: C? = holder1.item//编译错误

如果希望只插入，不取出，或者是方法的参数是泛型返回值没有就使用? super B
也叫做下边界通配符，逆变

      GenericHolder<? super B> holder2 = new GenericHolder<>();
        A a = new A();
        B b = new B();
        C c = new C();

        holder2.putItem(a);//编译错误
        holder2.putItem(b);
        holder2.putItem(c);


        A a1 = holder2.getItem();//编译错误
        B b1 = holder2.getItem();//编译错误
        C c1 = holder2.getItem();//编译错误

        val holder2:GenericHolder<in B> = GenericHolder()
        val a = A()
        val b = B()
        val c = C()
         holder2.putItem(a)/编译错误
        holder2.putItem(b)
        holder2.putItem(c)

        val a1: A? = holder1.item  //编译错误
        val b1: B? = holder1.item / /编译错误
        val c1: C? = holder1.item //编译错误

补充
kotlin中的*
MutableList<*> 表示的是 MutableList<out Any?>
Kotlin 中可以用UnsafeVariance注解打破编译器对in out的限制但同时也会失去运行时的状态确认，参考 collections.kt

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