C ++：如何要求一个模板类型派生自另一个 [英] C++: How to require that one template type is derived from the other

问题描述

在比较运算符中：

template<class R1, class R2>
bool operator==(Manager<R1> m1, Manager<R2> m2) {
    return m1.internal_field == m2.internal_field;
}

有没有办法我可以强制R1和R2必须有一个超类型子类型关系？也就是说，我想允许R1从R2派生，或者R2从R1派生，但是如果R1和R2是不相关的类型，则不允许比较。

Is there any way I could enforce that R1 and R2 must have a supertype or subtype relation? That is, I'd like to allow either R1 to be derived from R2, or R2 to be derived from R1, but disallow the comparison if R1 and R2 are unrelated types.

推荐答案

您想要的特征可能如下所示：

A trait you want might look like this:

template <typename B, typename D>
struct is_base_of // check if B is a base of D
{
    typedef char yes[1];
    typedef char no[2];

    static yes& test(B*);
    static no& test(...);

    static D* get(void);

    static const bool value = sizeof(test(get()) == sizeof(yes);
};

那么你只需要一个静态的断言：

Then you just need a static assert of some sort:

// really basic
template <bool>
struct static_assert;

template <>
struct static_assert<true> {}; // only true is defined

#define STATIC_ASSERT(x) static_assert<(x)>()

然后将两者放在一起：

template<class R1, class R2>
bool operator==(Manager<R1> m1, Manager<R2> m2)
{
    STATIC_ASSERT(is_base_of<R1, R2>::value || is_base_of<R2, R1>::value);

    return p1.internal_field == p2.internal_field;
}

如果没有派生（你的错误将类似于 static_assert< false> 未定义，它将指向该行。）

If one does not derive from the other, the function will not compile. (Your error will be similar to "static_assert<false> not defined", and it will point to that line.)

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