c++ - 正在访问指针元组的元组和线程安全的互斥锁

Question

给定std::tuple，

using Tuple1 = std::tuple<Foo1*, Bar1*, std::shared_ptr<std::mutex>>;
using Tuple2 = std::tuple<Foo2*, Bar2*, std::shared_ptr<std::mutex>>;
std::tuple<Tuple1, Tuple2> tuple;

还有函数，

void baz()
{
    auto tup = std::get<0>(tuple);

    std::lock_guard<std::mutex> lk(*std::get<2>(tup));

    // Do something with std::get<0>(tup) and std::get<1>(tup)
}

根据 this 关于 SO 访问 std::tuple 的问题本质上不是线程安全的，但是在示例代码的情况下呢？是否有可能发生未定义/奇怪的事情？

这是假设 FooN 和 BarN 仅在锁定后才被访问。

Answer 1

引用您所链接问题的完美答案:

However, if the parameter were const, then get would not be considered to provoke a data race with other const calls to get.

这基本上就是您的答案。 对 const 元组进行每次 get 调用(在任何未完全受互斥锁保护的元组上)，您安全。

这意味着您发布的代码不安全。像这样修改:

void baz()
{
    //    vvvv just being explicit here
    auto const & tup = std::get<0>(static_cast<decltype(tuple) const &>(tuple));

    std::lock_guard<std::mutex> lk(*std::get<2>(tup));

    // Dereference std::get<0>(tup) and std::get<1>(tup), 
    // use the pointed to objects at will, nothing else

    // Not ok, because it could interfer with the call in initialisation of tup of another thread
    // auto non_const_tup = std::get<0>(tuple)
}

目前我看到的唯一解决方案是使用像这样的元组:

std::tuple<
    std::shared_pointer<std::mutex>,
    std::unique_pointer<std::tuple<Foo1*, Bar1*>>
    // Mutex and pointer to tuple for Foo2 and Bar2
    >

所需的 const 将坚持一切(指针目标除外)。