正在铸造 std::pair<T1, T2>常量&到 std::pair&l

问题描述

reinterpret_cast std::pair 是否安全(理论上或实践中)?const & 转换为 std::pairconst &，假设程序员没有故意做一些奇怪的事情，比如专门化 std::pair?

Is it safe (in theory or in practice) to reinterpret_cast a std::pair<T1, T2> const & into a std::pair<T1 const, T2> const &, assuming that the programmer hasn't intentionally done something weird like specializing std::pair<T1 const, T2>?

推荐答案

这样做不可移植.

std::pair 要求在第 20.3 条中列出.第 17.5.2.3 条阐明

std::pair requirements are laid out in clause 20.3. Clause 17.5.2.3 clarifies that

第 18 条到第 30 条和附件 D 没有指定类的表示，并有意省略了类成员的说明.一个实现可以根据需要定义静态或非静态类成员，或两者都定义，以实现第 18 至 30 条和附件 D 中指定的成员函数的语义.

Clauses 18 through 30 and Annex D do not specify the representation of classes, and intentionally omit specification of class members. An implementation may define static or non-static class members, or both, as needed to implement the semantics of the member functions specified in Clauses 18 through 30 and Annex D.

这意味着实现包含部分特化是合法的(尽管不太可能)，例如:

This implies that it's legal (although incredibly unlikely) for an implementation to include a partial specialization such as:

template<typename T1, typename T2>
struct pair<T1, T2>
{
    T1 first;
    T2 second;
};

template<typename T1, typename T2>
struct pair<const T1, T2>
{
    T2 second;
    const T1 first;
};

显然与布局不兼容.该规则还允许其他变体，包括可能在 first 和/或 second 之前包含额外的非静态数据成员.

which are clearly not layout-compatible. Other variations including inclusion of additional non-static data members possibly before first and/or second are also allowed under the rule.

现在，考虑布局已知的情况有点有趣.尽管 Potatoswatter 指出 DR1334 断言 T 和 const T 不是 layout-compatible，标准提供了足够的保证，让我们无论如何都能获得大部分方式:

Now, it is somewhat interesting to consider the case where the layout is known. Although Potatoswatter pointed out DR1334 which asserts that T and const T are not layout-compatible, the Standard provides enough guarantees to allow us to get most of the way anyway:

template<typename T1, typename T2>
struct mypair<T1, T2>
{
    T1 first;
    T2 second;
};

mypair<int, double> pair1;
mypair<int, double>* p1 = &pair1;
int* p2 = reinterpret_cast<int*>(p1); // legal by 9.2p20
const int* p3 = p2;
mypair<const int, double>* p4 = reinterpret_cast<mypair<const int, double>*>(p3); // again 9.2p20

然而这不适用于 std::pair 因为我们不能在不知道 first 实际上是未指定的初始成员的情况下应用 9.2p20.

However this doesn't work on std::pair as we can't apply 9.2p20 without knowing that first is actually the initial member, which is not specified.

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