确保通用lambda表达式的类型安全性

Question

这有点是Java8从类中检索lambda setter的后续工作.

我正在尝试为给定字段获取getter方法

public <T, R> IGetter<T, R> getGetter(Class<T> clazz, Field field) {

    Class<R> fieldType = null;
    try {
        fieldType = (Class<R>) field.getType();
    } catch(ClassCastException e) {
        error("Attempted to create a mistyped getter for the field " + field + "!");
    }

    return getGetter(clazz, field.getName(), fieldType);
}

这是基本方法:

public <T, R> IGetter<T, R> getGetter(Class<T> clazz, String fieldName, Class<R> fieldType) {

    MethodHandles.Lookup caller = null;
    MethodHandle target = null;
    MethodType func = null;

    try {
        caller = MethodHandles.lookup();
        MethodType getter = MethodType.methodType(fieldType);
        target = caller.findVirtual(clazz, computeGetterName(fieldName), getter);
        func = target.type();
    } catch (NoSuchMethodException e) {
        error("Could not locate a properly named getter \"" + computeGetterName(fieldName) + "\"!");
    } catch (IllegalAccessException e) {
        error("Could not access \"" + computeGetterName(fieldName) + "\"!");
    }

    CallSite site = null;
    try {
        site = LambdaMetafactory.metafactory(
                caller,
                "get",
                MethodType.methodType(IGetter.class),
                func.generic(),
                target,
                func
        );
    } catch (LambdaConversionException e) {
        error("Could not convert the getter \"" + computeGetterName(fieldName) + "\" into a lambda expression!");
    }

    MethodHandle factory = site.getTarget();

    IGetter<T, R> r = null;
    try {
        r = (IGetter<T, R>) factory.invoke();
    } catch (Throwable throwable) {
        error("Casting the factory of \"" + computeGetterName(fieldName) + "\" failed!");
    }

    return r;
}

由于类型不匹配,这不会编译:

IGetter<TestEntity, Long> getter = accessorFactory.getGetter(TestEntity.class, "name", String.class);

但是这会编译:

Field field = TestEntity.class.getDeclaredField("name");
IGetter<TestEntity, Long> getter = accessorFactory.getGetter(TestEntity.class, field);

而且,令我惊讶的是,这确实可以使用上面检索到的getter:

TestEntity testEntity = new TestEntity(1L, "Test");
System.out.println(getter.get(testEntity));

但是,一旦我这样做:

Long value = getter.get(testEntity);

我得到以下异常:

Exception in thread "main" java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Long
    at de.cyclonit.exercise.Main.main(Main.java:26)

有没有办法早点抓住这个？

TestEntity类:

public class TestEntity {

    private Long id;

    private String name;


    public TestEntity(Long id, String name) {
    this.id = id;
        this.name = name;
    }


    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }
}

Answer 1

问题是你的方法

public <T, R> IGetter<T, R> getGetter(Class<T> clazz, Field field) {
    Class<R> fieldType = null;
    try {
        fieldType = (Class<R>) field.getType();
    } catch(ClassCastException e) {
        error("Attempted to create a mistyped getter for the field " + field + "!");
    }
    return getGetter(clazz, field.getName(), fieldType);
}

实际上并没有进行检查.您基本上是将Class实例转换为Class无效的类型.泛型类型的变化Class<?>来Class<R>是一个纯粹的编译时的事情,这就是为什么你应该在此时得到一个"未登记"的警告,至少,如果所有的警告被启用.

在运行时进行实际检查的唯一方法是从调用者获取预期的类型:

public <T, R> IGetter<T, R> getGetter(Class<T> clazz, Class<R> fieldType, Field field) {
    if(fieldType != field.getType()) {
        error("Attempted to create a mistyped getter for the field " + field + "!");
    }
    return getGetter(clazz, field.getName(), fieldType);
}

当然,它使得该方法接受Field有点无意义.实际上getGetter已经执行了一个需要与getter的返回类型完全匹配的查找,并且你不需要获得字段的类型和getter的返回类型必须匹配.实际上,它会对内部细节产生不必要的依赖.

为什么不简单地注释getter而不是字段......