Question

我有这个设计，我不确定它为什么不起作用。

interface BaseType {}

interface TypeA extends BaseType {}

interface TypeB extends BaseType {}

interface Query<T extends BaseType> {
    public String get();
}

interface Result<T extends BaseType> {
    public String get();
}

interface Service<T extends BaseType> {
    public Result<T> get(Query<T> query);
}

class SomeResult implements Result<TypeA> {
    private String s;
    public SomeResult(String s) { this.s = s; }
    public String get() { return this.s; }
}

class SomeQuery implements Query<TypeA> {
    public String get() { return "blah"; }
}

class SomeQuery2 implements Query<TypeA> {
    public String get() { return "blah2"; }
}

class SomeService implements Service<TypeA> {
    /** OK -- but notice the ambiguous parameter type */
    /*
    public SomeResult get(Query<TypeA> query) {
        if (query instanceof SomeQuery) return new SomeResult(query.get());
        else return null;
    }
    */

    /** NOT OK -- but this is the parameter I want to keep; notice SomeQuery IS-A Query<TypeA> */
    public SomeResult get(SomeQuery query) { return new SomeResult(query.get()); };
    /**
     * Main.java:27: error: SomeService is not abstract and does not override abstract method get(Query<TypeA>) in Service
     * class SomeService implements Service<TypeA> {
     * ^
     * 1 error
     */
}

public class Main {
    public static void main(String args[]) {
        SomeQuery someQuery = new SomeQuery();
        SomeQuery2 someQuery2 = new SomeQuery2();
        SomeService someService = new SomeService();
        System.out.println(someService.get(someQuery).get());
    }
}

我是仿制药的新手，并不太明白我在这里违反了什么合同。我希望服务紧密有限，即使我可以绑定返回类型，我似乎也不能这样做参数。这意味着，我需要在服务中进行instanceof检查，以确保我获得正确的参数。我想避免这种情况。有什么想法吗？

Answer 1

由于return type covariance，您可以使覆盖方法的返回类型更具体，但是您无法在不更改其签名的情况下更改方法的参数。这就是编译器抱怨您在将其更改为get(Query<TypeA>)时未实现get(SomeQuery)的原因。您需要使Service更灵活，才能获得所需内容：

interface Service<T extends BaseType, Q extends Query<T>> {
    public Result<T> get(Q query);
}

class SomeService implements Service<TypeA, SomeQuery> {

    @Override
    public SomeResult get(SomeQuery query) {
        ...
    }
}

另请注意，编码到界面时缩小的返回类型无关紧要：当SomeService键入为Service<TypeA, SomeQuery>时，get仍会返回Result<TypeA>。因此，您可以考虑对结果类型进行类似的更改：

interface Service<T extends BaseType, Q extends Query<T>, R extends Result<T>> {
    public R get(Q query);
}

class SomeService implements Service<TypeA, SomeQuery, SomeResult> {

    @Override
    public SomeResult get(SomeQuery query) {
        ...
    }
}

泛型有界类型参数

1 个答案: