拒绝std :: vector删除其数据 [英] Deny std::vector from deleting its data

问题描述

我有以下情况：

T* get_somthing(){
    std::vector<T> vec; //T is trivally-copyable
    //fill vec
    T* temp = new T[vec.size()];
    memcpy(temp, vec.data(), vec.size() * sizeof(T));
    return temp;
}

我想通过返回<$ c $来摆脱复制过程c> std :: vector :: data 直接像这样：

I want to get rid of the copy process by returning the std::vector::data directly like this:

T* get_somthing(){
    std::vector<T> vec; //T is trivally-copyable
    //fill vec
    return temp.data();
}

但是，这是错误的，因为当<$时数据将被删除c $ c> vec 析构函数被调用。

However, that is wrong since the data is going to be deleted when vec destructor is called.

那么，如何防止vec删除其数据？换句话说，我想要某种从 std :: vector 到C ++ Raw Dynamic Array的移动习惯。

So, how can I prevent vec from delete its data? In other word I want some kind of move-idiiom from std::vector to C++ Raw Dynamic Array.

聚苯乙烯更改设计不是一种选择。使用 std :: vector 是强制性的。将指针返回到 array 也是强制性的。 Becauese它是两个模块之间的包装。一个需求向量，另一个需求指针。

P.S. Changing the design is not an option. Using the std::vector there is mandatory. Returning a pointer to array is also mandatory. Becauese It is a wrapper between two modules. One need vector the other need pointer.

推荐答案

PS。更改设计不是一种选择。使用std :: vector是强制性的。返回数组的指针也是必须的。

P.S. Changing the design is not an option. Using the std::vector there is mandatory. Returning a pointer to array is also mandatory.

更改设计是您的最佳选择。我建议重新考虑这种立场。

Changing the design is your best option. I recommend reconsidering this stance.

（目前^†）没有办法窃取向量的缓冲区，因此给定（问题中提到的限制（silly ^††），复制是可行的方法。

There is (currently^†) no way to "steal" the buffer of a vector, so given the (silly^††) limitations stated in the question, copying is the way to go.

†Tomasz Lewowski链接了一个提案，如果它包含在将来的标准中： http：//www.open- std.org/jtc1/sc22/wg21/docs/papers/2015/n4359.pdf （编辑：如前所述，它被c ++ 17拒绝了）

† Tomasz Lewowski linked a proposal that would change this if it is included in a future standard: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4359.pdf ( as pointed out, it was rejected from c++17)

††直到被具体要求证明是愚蠢的。

†† Silly until justified by concrete requirements.

这是一个包装在两个模块之间。一个需求向量，另一个需求指针。

It is a wrapper between two modules. One need vector the other need pointer.

大概，另一个需要指针的接口将缓冲区的破坏委托给调用者，可能使用了诸如 void delete_somthing（T *）之类的回叫。在我看来，不归还所有权将是非常糟糕的设计。

Presumably, the other interface that needs the pointer, delegates the destruction of the buffer to the caller, possibly using some sort of call back like void delete_somthing(T*). Taking ownership without giving it back would have been very bad design, in my opinion.

如果您确实控制了销毁，则可以将矢量存储在地图中，并在传递指针进行销毁时删除向量：

In case you do have control of the destruction, you can store the vector in a map, and erase the vector, when the pointer is passed for destruction:

std::unordered_map<T*, std::vector<T>> storage;

T* get_somthing(){
    std::vector<T> vec; //T is trivally-copyable
    //fill vec
    T* ptr = vec.data();
    storage[ptr] = std::move(vec);
    return ptr;
}

void delete_somthing(T* ptr){
    storage.erase(ptr);
}

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