std::vector 的 posix_memalign

发布于 2024-08-23 18:26:47 字数 206 浏览 4 评论 0原文

有没有办法在不先创建向量的本地实例的情况下对 std::vector 进行 posix_memalign ？我不知道该怎么说

sizeof(std::vector<type>(n))

我遇到的问题是我需要告诉 posix_memalign 要分配多少空间，但在没有实际创建新向量的情况下

。谢谢

原文

Is there a way to posix_memalign a std::vector without creating a local instance of the vector first?
The problem I'm encountering is that I need to tell posix_memalign how much space to allocate and I don't know how to say

sizeof(std::vector<type>(n))

without actually creating a new vector.

Thanks

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憧憬巴黎街头的黎明 2024-08-30 18:26:47

嗯，这里有两种尺寸。矢量本身通常只不过是一个或两个指向某些已分配内存的指针，以及跟踪大小和容量的无符号整数。还有分配的内存本身，这就是我认为你想要的。

您想要做的是创建一个向量将使用的自定义分配器。到时候，它将使用您的分配器，您可以拥有自己的特殊功能。我不会详细介绍分配器的完整详细信息，但具体细节如下：

template <typename T>
struct aligned_allocator
{
    // ...

    pointer allocate(size_type pCount, const_pointer = 0)
    { 
        pointer mem = 0;
        if (posix_memalign(&mem, YourAlignment, sizeof(T) * pCount) != 0)
        {
            throw std::bad_alloc(); // or something
        }

        return mem;
    }

    void deallocate(pointer pPtr, size_type)
    { 
        free(pPtr);
    }

    // ...
};

然后你可以像这样使用它：

typedef std::vector<T, aligned_allocator<T> > aligned_T_vector;

aligned_T_vector vec;
vec.push_back( /* ... */ ); // array is aligned

但是重申第一点，无论它包含多少个元素，向量的大小都是相同的，因为它只指向一个缓冲区。仅该缓冲区的大小发生变化。

Well, there are two sizes here. The vector itself is typically no more than a pointer or two to some allocated memory, and unsigned integers keeping track of size and capacity. There is also the allocated memory itself, which is what I think you want.

What you want to do is make a custom allocator that the vector will use. When it comes time, it will use your allocator and you can have your own special functionality. I won't go over the full details of an allocator, but the specifics:

template <typename T>
struct aligned_allocator
{
    // ...

    pointer allocate(size_type pCount, const_pointer = 0)
    { 
        pointer mem = 0;
        if (posix_memalign(&mem, YourAlignment, sizeof(T) * pCount) != 0)
        {
            throw std::bad_alloc(); // or something
        }

        return mem;
    }

    void deallocate(pointer pPtr, size_type)
    { 
        free(pPtr);
    }

    // ...
};

And then you'd use it like:

typedef std::vector<T, aligned_allocator<T> > aligned_T_vector;

aligned_T_vector vec;
vec.push_back( /* ... */ ); // array is aligned

But to reiterate the first point, the size of a vector is the same regardless of how many elements it's holding, as it only points to a buffer. Only the size of that buffer changes.

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