问题描述
如果我让 unique_ptr 指向一个 STL 容器的实例会怎样?这段代码安全吗?
What if I make a unique_ptr point to an instance of an STL container as follows? Is this code safe?
unique_ptr< vector<int> > p1( new vector<int> );
这不会导致 vector 的析构函数被调用两次，因为 vector 本身和 unique_ptr 都试图清理 vector 到目前为止获得的内存?这会导致未定义的行为吗?或者编译器是否知道 vector 已经释放了它的内存并且不会为了 unique_ptr 超出范围而再次调用析构函数?
Wouldn't this result in the destructor for the vector<int> being called twice, since both the vector<int> itself and the unique_ptr both attempt to clean up the memory the vector<int> had acquired so far? Could this result in undefined behavior? Or does the compiler somehow knows that that vector<int> has released its memory and does not invoke the destructor again for the sake of the unique_ptr going out of scope?
这只是为了理解，如果有人愚蠢到这样做，会不会很危险?
This is simply to understand that if someone was stupid enough to do this, could it be dangerous?
推荐答案
With unique_ptr<向量>p1( new vector ); unique_ptr 在 vector 上调用 delete.vector 的析构函数将释放自己分配的内存.所以是安全的.
With unique_ptr< vector<int> > p1( new vector<int> ); the unique_ptr with call delete on the vector.The destructor of vector will then release its own allocated memory.So it is safe.
但是 vector 就足够了.我没有看到您想要 unique_ptr< 的情况.向量>.
But vector<int> is enough. I don't see a case where you'd want unique_ptr< vector<int> >.
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