std :: unique_ptr< T []>和自定义分配器删除器 [英] std::unique_ptr<T[]> and custom allocator deleter

问题描述

我试图使用自定义内存分配器 std :: unique_ptr< T []> 。基本上，我有自定义分配器，它是 IAllocator 的子类，它提供了以下方法：

I am trying to use std::unique_ptr<T[]> with custom memory allocators. Basically, I have custom allocators that are subclasses of IAllocator, which provides the following methods:

void* Alloc( size_t size )
template<typename T> T* AllocArray( size_t count )
void Free( void* mem )
template<typename T> void FreeArray( T* arr, size_t count )

由于底层内存可能来自预分配块，我需要特殊的 ... Array（） -methods来分配和释放数组，他们分配/释放内存并调用 T（） / code> / 〜T（）。
现在，据我所知， std :: unique_ptr 的自定义删除器使用签名：

Since the underlying memory might come from a pre-allocated block, I need the special ...Array()-methods to allocate and free arrays, they allocate/free memory and call T() / ~T() on every element in the range. Now, as far as I know, custom deleters for std::unique_ptr use the signature:

void operator()(T* ptr) const

unique_ptr< T []> 的情况下，通常你会调用 delete [] 但我必须调用 FreeArray< T> ，我需要的范围内的元素数量。只有原始的指针，我想没有办法获得范围的大小，因此我唯一能想出的是这样：

In the case of unique_ptr<T[]>, normally you would call delete[] and be done with it, but I have to call FreeArray<T>, for which I need the number of elements in the range. Given only the raw pointer, I think there is no way of obtaining the size of the range, hence the only thing I could come up with is this:

std::unique_ptr<T[], MyArrDeleter> somePtr( allocator.AllocArray<T>( 20 ), MyArrDeleter( allocator, 20 ) );

其中数组的大小基本上必须手动传递到删除对象中。有没有更好的方法来做到这一点？这似乎对我来说很容易出错...

Where essentially the size of the array has to be passed into the deleter object manually. Is there a better way to do this? This seems quite error-prone to me...

推荐答案

是的，最肯定有一个更好的方法：

使用maker-function。

Yes, there most certainly is a better way:
Use a maker-function.

template<class T, class A> std::unique_ptr<T[], MyArrDeleter>
my_maker(size_t count, A&& allocator) {
    return {somePtr(allocator.AllocArray<T>(count), MyArrDeleter(allocator, count)};
}

auto p = my_maker<T>(42, allocator);

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