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我正在尝试在 java 中定义一些基本的数学概念，但总是遇到泛型错误。考虑下面的例子。在此示例中，我尝试将零映射（所有 x 的 f(x) = 0）定义为常量映射（所有 x 的 f(x) = c）和线性映射（f(a*x + b*y) = a*f(x) + b*f(y))。编译器不允许我这样做，因为“零映射是一个向量”这句话是模棱两可的（在 java 方面，但不是在数学方面）。

你能建议一个干净的解决方案来解决这个问题吗？

// A vector in a finite-dimensional vector space over the real numbers.
interface Vector<V extends Vector<?>>
{
    int dimension();
    V plus(V v);
    V times(double c);
}

interface Mapping<U extends Vector<?>, V extends Vector<?>>
// Does not inherit from Vector because the set of all mappings is an
// infinite-dimensional vector space.
{
    V map(U u);
}

// Linear mappings, on the other hand, from one finite-dimensional vector space
// to another, do form a finite-dimensional vector space.
interface LinearMapping<U extends Vector<?>, V extends Vector<?>>
    extends Mapping<U, V>, Vector<LinearMapping<U, V>>
{
}

// All elements of U are mapped to getImage(). If V is finite-dimensional, then
// the set of constant mappings is also a finite-dimensional vector space.
interface ConstMapping<U extends Vector<?>, V extends Vector<?>>
    extends Mapping<U, V>, Vector<ConstMapping<U, V>>
{
    V getImage();
}

// A zero mapping is both constant and linear, but cannot be defined as such.
interface ZeroMapping<U extends Vector<?>, V extends Vector<?>>
    extends LinearMapping<U, V>, ConstMapping<U, V>
// Error: The interface Vector cannot be implemented more than once with
// different arguments: Vector<ConstMapping<U,V>> and Vector<LinearMapping<U,V>>
{
}
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1 回答 1

3

问题似乎是 ZeroMapping 正在扩展Vector<LinearMapping>和Vector<ConstMapping>.

为了解决这个问题，我们将泛型参数 VLinearMapping和更改ConstMapping为其各个类的下限，以便它们可以扩展泛型Vector<V>，从而提供更大的灵活性。希望这能捕捉到您将 LinearMapping 制作为 Mapping 和 Vector（仅其本身）的意图。

编辑：在这种情况下，您可能必须使用 3 个通用参数：2 捕获“映射”，1 捕获与其他相同类型的线性映射的“向量”关系。

ZeroMapping 然后遵循相同的格式。

我在下面的回复中包含了@LuiggiMendoza 的建议。

interface Vector<V extends Vector<V>>
{
    int dimension();
    V plus(V v);
    V times(double c);
}

interface Mapping<U extends Vector<U>, V extends Vector<V>>
// Does not inherit from Vector because the set of all mappings is an
// infinite-dimensional vector space.
{
    V map(U u);
}

// Linear mappings, on the other hand, from one finite-dimensional vector space
// to another, do form a finite-dimenszional vector space.
interface LinearMapping<U extends Vector<U>, V extends Vector<V>, W extends LinearMapping<U, V, W>>
    extends Mapping<U, V>, Vector<W>
{
}

// All elements of U are mapped to getImage(). If V is finite-dimensional, then
// the set of constant mappings is also a finite-dimensional vector space.
interface ConstMapping<U extends Vector<U>, V extends Vector<V>, W extends ConstMapping<U, V, W>>
    extends Mapping<U, V>, Vector<W>
{
    V getImage();
}

// A zero mapping is both constant and linear, but cannot be defined as such.
interface ZeroMapping<U extends Vector<U>, V extends Vector<V>, W extends ZeroMapping<U, V, W>>
    extends LinearMapping<U, V, W>, ConstMapping<U, V, W>
{
}

如何实现该类的一个示例可能是：

class RealVector implements Vector<RealVector> { // methods here .. // }

class RealLinearMapping implements LinearMapping<RealVector, RealVector, RealLinearMapping>
{
    @Override
    public RealVector map(RealVector u) { ... }

    @Override
    public RealLinearMapping plus(RealLinearMapping v) { ... }
}
于 2013-05-26T14:08:00.067 回答