c++ std::lock_guard 作用域到达

Question

假设我有一个获取锁并执行由参数传递的函数的函数：

template <typename T>
T acquireLock(std::mutex& _lock, const std::function<T()>& execution) {

  try {
    std::lock_guard<std::mutex> mutex (_lock);

    return execution();

  } catch (std::logic_error& error) {

    std::cerr << "[exception caught]\n\t" << error.what() << std::endl;

  }

  return false;
}

此外，我有一个 class 需要为其某些方法获取所述锁。

class MyThreadSafeClass {

 public:

  bool Init();
  bool StopApi();
  unsigned int GetValue() {

      auto ret = acquireLock<unsigned int>(_lock, [this]() -> unsigned int {

        // does some work that's not thread-safe...
        return value;

      });

      return ret;
  }

 private:

  bool _ready = false;
  std::mutex _lock;

};

我怀疑每当我调用 GetValue() 时，看看我的 acquireLock() 方法，execution() 调用是否也受锁定范围的影响？

auto myClass = new MyThreadSafeClass();
myClass->GetValue();

查看 this，更具体地说：

When a lock_guard object is created, it attempts to take ownership of the mutex it is given. When control leaves the scope in which the lock_guard object was created, the lock_guard is destructed and the mutex is released.

我仍然不清楚 execution() 代码内部发生的事情是否仍然受锁定范围的影响。

Answer 1

根据[stmt.return]/p3：

The copy-initialization of the result of the call is sequenced before the destruction of temporaries at the end of the full-expression established by the operand of the return statement, which, in turn, is sequenced before the destruction of local variables ([stmt.jump]) of the block enclosing the return statement.

所以我们得到：

互斥量已锁定
execution() 在持有锁时被调用
锁被释放
计算值返回给调用者（或输入catch子句）

换句话说，是的，它将按预期工作。

无关说明：std::function效率不高。对可调用类型进行模板化应该会更好：

template<typename F>
auto doLocked(std::mutex& _lock, F const& f) -> decltype(f()) {
    std::lock_guard<std::mutex> lock(_lock);
    return f();
}

c++ std::lock_guard 作用域到达

c++ std::lock_guard scope reach

c++

mutex

locks