c++ - 从 const 成员函数返回的对象的常量性

Question

关于 const 成员函数的返回值是否应为 const 的经验法则是什么？这是我倾向于做的，但我挣扎于模棱两可的场景。

class foo
{
    public:
        // if the returned object is conceptually
        // (and therefore usually literally but not even necessarily) 
        // owned by *this*,
        // then a const ptr should be returned
        const ownedByFoo *getOwnedByFoo() const { return m_ownedByFoo; }
        ownedByFoo *getOwnedByFoo() { return m_ownedByFoo; }

        // what's a poor boy to do here?
        // veryRelatedToFoo might be a conceptual contemporary or parent
        // of Foo. Note naming the function "find..." instead of "get.."
        // since *get* implies the object holds it as a data member but
        // that may not even be the case here. There could be a
        // standalone container that associates foo and veryRelatedToFoo.
        // Should constness be symmetrical here?
        const veryRelatedToFoo *findVeryRelatedToFoo() const;
        veryRelatedToFoo *findVeryRelatedToFoo();

        // If the returned object is only (conceptually)
        // peripherally related to *this*, 
        // the returned object probably shoudn't be const, even if
        // foo happens to have one of these as a direct data member,
        // since we don't want to give away the implementation details of
        // Foo and it may hold peripherallyRelatedToFoo simply for performance
        // reasons, etc.
        peripherallyRelatedToFoo *findPeripherallyRelatedToFoo() const

    ...
};

另外要注意的是，一旦你有了非对称常量，你就可以让 const 对象 A 返回对象 B，然后让对象 B 返回对象 A，然后那么你已经设法绕过了对象 A 的预期常量性。

Answer 1

拥有任何常量比你在许多语言中所能找到的都要多!

一旦您理解(正如您所理解的那样)“物理常量”和“逻辑常量”之间的区别，那么“模棱两可”的场景就仍然存在:有待商榷。例如考虑 mutable 关键字存在的事实。

我认为您的总体想法是正确的:除此之外，所有这些都是有争议的和主观的，并且会根据特定问题和约束等而有所不同。

我倾向于“没有异常(exception)”，所以我会避免从 const 方法返回指向非常量的指针；但我想我可以想象一些我可能会被诱惑的场景……尽管我可能更有可能决定使该方法成为非常量。