问题描述
请考虑以下内容:
std::vector<std::unique_ptr<int>> ptrsToInts;
ptrsToInts.emplace_back(new int);
如果向量中发生了重新分配，但失败了(抛出std::bad_alloc)，我是安全的"还是泄漏int?
If reallocation occurs in the vector, and that fails (throwing std::bad_alloc), am I "safe" or will I leak an int?
C ++ 11 23.3.6.5 [vector.modifiers]/1说:
C++11 23.3.6.5 [vector.modifiers]/1 says:
似乎表明这是一个潜在问题.也就是说，如果没有效果"，那么就不会构造unique_ptr，因此析构函数的行为将依赖于delete不会发生指针. (这可能表明emplace_back的容器应禁止使用emplace_back)
which seems to indicate that this is a potential problem. That is, if there are "no effects", then no unique_ptr ever was constructed, and therefore the destructor behavior one would rely on to delete that pointer would not occur. (Which might indicate that emplace_back should be banned for containers of unique_ptrs)
推荐答案
如果需要重新分配并且失败，那么是的，您的对象永远不会进入容器，因此会丢失.
If reallocation is required and it fails, then yes, your object never went into the container and will thus be lost.
但是，应注意，这是纯用户错误.对于unique_ptr的容器，不应禁止" emplace_back，因为这样做有非常安全的方式(例如，事先reserve留出空间，因此您知道它将一直存在).另外，您可以传入整个unique_ptr，因为它完全可以使用move构造函数.
However, it should be noted that this is pure user error. emplace_back should not be "banned" for containers of unique_ptr, because there are perfectly safe ways of doing this (such as reserveing the space beforehand, so you know it will always be there). Also, you can pass in whole unique_ptrs, since it's perfectly capable of using a move constructor.
所以，确实，在抛出异常之前，没有正确地将非RAII对象(int*)包装在RAII对象中是您的错.
So really, it's your fault for not properly wrapping your non-RAII object (the int*) in a RAII object before the point where you could throw exceptions.
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