从std :: vector接管内存 [英] taking over memory from std::vector

问题描述

我使用一个处理大量数据的外部库。数据由原始指针加上长度传递。该库不声明指针的所有权，而是在处理完数据后调用提供的回调函数（具有两个相同的参数）。

数据已准备就绪通过使用 std :: vector< T> 方便，我宁愿不放弃这种便利。复制数据完全是不可能的。因此，我需要一种方法来接管 std :: vector< T> 所拥有的内存缓冲区，然后在回调中将其释放。 / p>

我当前的解决方案如下：

  std :: vector< &输入= prepare_input（）; 
 T *数据= input.data（）; 
 size_t size = input.size（）; 
 //将向量移至原始存储区，以防止释放
 alignas（std :: vector< T>）char temp [sizeof（std :: vector< T>））]; 
 new（temp）std :: vector< t（std :: move（input））; 
 //调用库
 lib :: startProcesing（data，size）; 
  

，然后在回调函数中：

  void回调（T * data，size_t size）{
 std :: allocator< T>（）。deallocate（data，size）; 
} 
  

此解决方案有效，因为标准分配器的 deallocate 函数将忽略其第二个参数（元素计数），仅调用 :: operator delete（data）。否则，可能会发生不好的事情，因为输入向量的大小可能比其容量。

我的问题是：是否有一种可靠的（写成C ++标准）方式来接管 std :: vector的缓冲区并在以后的某个时间手动发布？

解决方案

您不能拥有此所有权向量中的内存，但是您可以用另一种方式解决您的潜在问题。

这里是我的处理方式-由于静态全局变量和不是线程安全的，但是可以通过对注册表对象的访问进行一些简单的锁定来实现。

 静态std :: map< T *，std :: vector< T> *>注册表
 void my_startProcessing（std :: vector< T> *数据）{
 Registry.put（data-> data（），data）; 
 lib :: startProcesing（data-> data（），data-> size（））; 
} 
 
 void my_callback（T * data，size_t length）{
 std :: vector< T> *原始= Registry.get（数据）; 
删除原件； 
registry.remove（数据）; 
} 
  

现在您可以做

  std :: vector< T> *输入= ... 
 my_startProcessing（输入）; 
  

但是要小心！如果添加/删除元素，将会发生不好的事情调用 my_startProcessing 后输入的内容-库的缓冲区可能无效。 （您可能被允许更改向量中的值，因为我相信它将正确地写入数据，但是这也取决于库允许的内容。）

如果 T = bool 也是无效的，因为 std :: vector< bool> ; :: data（）不起作用。

I use an external library which operates on large quantities of data. The data is passed in by a raw pointer, plus the length. The library does not claim ownership of the pointer, but invokes a provided callback function (with the same two arguments) when it is done with the data.

The data gets prepared conveniently by using std::vector<T>, and I'd rather not give up this convenience. Copying the data is completely out of the question. Thus, I need a way to "take over" the memory buffer owned by an std::vector<T>, and (later on) deallocate it in the callback.

My current solution looks as follows:

std::vector<T> input = prepare_input();
T * data = input.data();
size_t size = input.size();
// move the vector to "raw" storage, to prevent deallocation
alignas(std::vector<T>) char temp[sizeof(std::vector<T>)];
new (temp) std::vector<T>(std::move(input));
// invoke the library
lib::startProcesing(data, size);

and, in the callback function:

void callback(T * data, size_t size) {
    std::allocator<T>().deallocate(data, size);
}

This solution works, because the standard allocator's deallocate function ignores its second argument (the element count) and simply calls ::operator delete(data). If it did not, bad things could happen, as the size of the input vector might be quite a bit smaller than its capacity.

My question is: is there a reliable (wrt. the C++ standard) way of taking over the buffer of std::vector and releasing it "manually" at some later time?

解决方案

You can't take ownership of the memory from a vector, but you can solve your underlying problem another way.

Here's how I'd approach it - its a bit hacky because of the static global variable and not thread safe, but it can be made so with some simple locking around accesses to the registry object.

static std::map<T*, std::vector<T>*> registry;
void my_startProcessing(std::vector<T> * data) {
  registry.put(data->data(), data);
  lib::startProcesing(data->data(), data->size());
}

void my_callback(T * data, size_t length) {
  std::vector<T> * original = registry.get(data);
  delete original;
  registry.remove(data);
}

Now you can just do

std::vector<T> * input = ...
my_startProcessing(input);

But watch out! Bad things will happen if you add/remove elements to the input after you've called my_startProcessing - the buffer the library has may be invalidated. (You may be allowed to change values in the vector, as I believe that will write through the to data correctly, but that will depend on what the library allows too.)

Also this doesn't work if T=bool since std::vector<bool>::data() doesn't work.

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