这个函数（运算符重载）是线程安全的吗？
这是代码：
ElementType& operator[] (int key)
{
//something like boost::mutex::scoped_lock
MutexLockType lock();
if(key < 0 || key > m_bound)
throw std::range_error("access out of bound");
return m_elements[key];
}
Here's the code:
ElementType& operator[] (int key)
{
//something like boost::mutex::scoped_lock
MutexLockType lock();
if(key < 0 || key > m_bound)
throw std::range_error("access out of bound");
return m_elements[key];
}
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不，这不是因为您允许元素的引用从函数中泄漏并远离锁的安全性。
更重要的是，如果没有更严格的规范，线程安全问题就有点难以回答 您最想要哪种类型的线程安全。至少您需要向我们展示
m_elements和m_bound的所有其他访问权限。No it is not because you have allowed a reference the element to leak out of the function and away from the safety of the lock.
What's more, thread-safety questions are a little hard to answer without a harder specification of just what flavour of thread-safety you are desirest. At the very least you would need to show us every other access of
m_elementsandm_bound.在这个确切的示例中，当您意识到根本没有锁，只有返回 LockType 的 lock() 函数的声明时，您会更加惊讶。
无论如何，这并不是说它会对锁有帮助。
In this exact example you would be even more surprised when you realize that there is no lock at all, just a declaration of a lock() function that returns a LockType.
Not that that it would have helped with a lock anyway.