java - 在@FunctionalInterface 中处理可变参数的更简洁方法？

Question

考虑这个接口(interface):

@FunctionalInterface
public interface ThrowingFunction<T, R>
{
    R tryApply(T t)
        throws Throwable;
}

这个类:

public final class Wrapper<T, R>
{
    private final ThrowingFunction<T, R> f;

    public Wrapper(final ThrowingFunction<T, R> f)
    {
        this.f = f;
    }
}

有了这些，我可以写:

final ThrowingFunction<Path, Path> f = Path::toRealPath;

new Wrapper(f);

但我不能写:

new Wrapper(Path::toRealPath);

问题是由于the method prototype .它可以将 LinkOption... 作为参数。

然而，编译器能够“强制”此处的方法引用。我想出的唯一模仿这个的方法是彻头彻尾的骇人听闻:

@FunctionalInterface
public interface ThrowingFunction<T, R>
{
    R tryApply(T t)
        throws Throwable;

    @FunctionalInterface
    interface WithVarags<T, V, R>
        extends ThrowingFunction<T, R>
    {
        R tryApplyWithVarargs(T t, V... ignored)
            throws Throwable;

        @Override
        default R tryApply(T t)
            throws Throwable
        {
            return tryApplyWithVarargs(t);
        }
    }
}

然后向Wrapper 添加一个新的构造函数:

public <V> Wrapper(
    final ThrowingFunction.WithVarags<T, V, R> function)
{
    this.function = function;
}

有没有更简洁的方法来做到这一点？也就是说，在这种情况下我可以“像编译器一样”吗？

Answer 1

您可以在实例化时指定 Wrapper 的类型。

new Wrapper<Path, Path>(Path::toRealPath);

或

Wrapper<Path, Path> wrapper = new Wrapper<>(Path::toRealPath);

我认为这与可变参数无关。通过指定类型，您还告诉编译器“忽略”varargs 参数。

---

请注意，我看到您也可以“忽略”同一函数的返回类型:

ThrowingBiConsumer<Path, LinkOption[]> biConsumer =  Path::toRealPath;
new Wrapper<Path, LinkOption[]>(Path::toRealPath);


public interface ThrowingBiConsumer<T, U> {
    void accept(T t, U u) throws Throwable;
}