我可以强制 std::vector 留下内存泄漏吗?

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在向量超出范围后，我可以强制 std::vector 不释放其内存吗?

Can I force std::vector to not deallocate its memory after the vector goes out of scope?

例如，如果我有

int* foo() { std::vector<int> v(10,1); // trivial vector return &v[0]; } int main() { int* bar = foo(); std::cout << bar[5] << std::endl; }

无法保证此处仍可访问这些值.

There is no guarantee that the values will still be accessible here.

我目前只是这样做

int* foo() { std::vector<int> v(10,1); int* w = new int[10]; for (int i=0; i<10; i++) { w[i] = v[i]; } return w; }

但是重新填充一个全新的数组有点浪费.有没有办法强制 std::vector 不删除它的数组?

but it is a little wasteful to repopulate a whole new array. Is there a way to force std::vector to not delete its array?

注意:我没有返回向量本身，因为我使用 SWIG 将 c++ 与 python 连接起来，并且 ARG_OUTVIEW_ARRAY 需要一个原始指针，实际上，这是一个有意的内存泄漏.然而，我仍然希望能够在构建数据本身的同时利用矢量特征.

Note: I am not returning the vector itself because I am interfacing c++ with python using SWIG, and ARG_OUTVIEW_ARRAY requires a raw pointer and, in fact, an intentional memory leak. I would still however like to be able to make use of vector features while constructing the data itself.

推荐答案

vector 旨在防止泄漏.

但如果你想用脚射击自己，这是可能的.以下是防止向量释放其内部数组的方法:

But if you want to shoot yourself in the foot, it's possible. Here's how you prevent the vector from deallocating its internal array:

int *foo() { std::vector<int> v(10,1); int *ret = v.data(); new (&v) std::vector<int>; // Replace `v` with an empty vector. Old storage is leaked. return ret; }

正如其他答案所说，你永远不应该这样做.

As the other answers say, you should never do it.