Question

在使用Kotlin / Java进行编码时，我在使用强制转换和泛型时偶然发现了一些奇怪的东西。似乎可以让类型系统认为列表属于List<Foo>类型，而它实际上是List<Object>。

任何人都可以向我解释为什么会这样吗？

以下是Kotlin和Java中的一个例子：

Kotlin的例子

fun <T> test(obj: Any): List<T> {
    val ts = ArrayList<T>()
    ts.add(obj as T)
    return ts
}

fun <T> test2(obj: Any): T {
    return obj as T
}

fun <T> test3(obj: Any): List<T> {
    val ts = ArrayList<T>()
    ts.add(test2(obj))
    return ts
}


fun main(args: Array<String>) {
    val x = test<Double>(1) // Returns a list of Integers and doesn't error
    println(x)

    val y = test2<Double>(1) // Casts the Int object to a Double.
    println(y)

    val z = test3<Double>(1) // Returns a list of Integers and doesn't error.
    println(z)
}

Java中的示例

public class Test {
    public static <T> List<T> test(Object obj){
        ArrayList<T> ts = new ArrayList<>();
        ts.add((T) obj);
        return ts;
    }

    public static <T> T test2(Object obj){
        return (T) obj;
    }

    public static <T> List<T> test3(Object obj){
        ArrayList<T> ts = new ArrayList<>();
        ts.add(test2(obj));
        return ts;
    }


    public static void main(String[] args) {
        List<Double> x = test(1); // Returns a list of Integers and doesn't error
        System.out.println(x);

        // Double y = test2(1); // Errors in java an Integers cannot be converted into a Double.
        // System.out.println(y);

        List<Double> z = test3(1); // Returns a list of Integers and doesn't error.
        System.out.println(z);
    }
}

Answer 1

Java没有具体化的泛型。也就是说，通用信息在运行时不存在，并且所有通用代码都是简化的＆＃34;通过一个称为擦除的过程。当已知泛型类型以确保正确性时，编译器会抛出强制转换。您不能转换为泛型类型，因为泛型类型不足以让运行时知道某个值是否为1，这就是javac对您大喊大叫的原因这样做，因为它知道你要求JVM做一些它不可能做的事情，导致运行时不安全。

public class Test {
    public static List test(Object obj) { // generic types => erasure = raw types
        ArrayList ts = new ArrayList();
        ts.add(obj); // No cast: List.add has erasure (Ljava.lang.Object;)V
        return ts;
    }

    public static Object test2(Object obj) { // T is unbounded => erasure = Object
        return obj; // No cast: all types <: Object
    }

    public static List test3(Object obj) {
        ArrayList ts = new ArrayList();
        ts.add(test2(obj)); // Note: we don't know what T is, so we can't cast to it and ensure test2 returned one.
        return ts;
    }


    public static void main(String[] args) {
        List x = test(1); // Returns a list and doesn't error
        System.out.println(x);

        Double y = (Double) test2(1); // Errors in java as an Integer cannot be converted into a Double
        // This is because the compiler needs to insert casts to make generics work
        System.out.println(y);

        List z = test3(1);
        // Unlike y, there isn't a cast in test3 because test3 doesn't know what T is, so the Integer passes through, uncast, into a List<Double>.
        // The JVM can't detect this, because it doesn't even know what a List<Double> is.
        System.out.println(z);
    }
}

请注意test2如何删除荣耀的身份函数，导致test3与test1完全相同，但具有间接级别。

Answer 2

编译器在执行未选中强制转换为T时发出警告。如果您的程序类型中出现此类警告，则无法保证安全。所以你看到的行为是预期的。

Answer 3

尝试更改第二个测试：

    Double y = test2(1.0);
    System.out.println(y);

在您的代码test2(1)中，参数1被自动生成为Integer，无法转换为Double。

Java＆amp; Kotlin铸造与仿制药。丢失类型安全

3 个答案: