C++的allocator
模板类本身并不直接提供对齐分配的功能,但是你可以通过自定义分配器来实现对齐分配。自定义分配器可以继承自std::allocator
并重载其allocate
和deallocate
方法,以便在分配和释放内存时考虑对齐要求。
以下是一个简单的示例,展示了如何创建一个支持对齐分配的自定义分配器:
#include <iostream>
#include <memory>
#include <vector>
template <typename T>
class aligning_allocator : public std::allocator<T> {
public:
using value_type = T;
using pointer = typename std::allocator<T>::pointer;
using const_pointer = typename std::allocator<T>::const_pointer;
using reference = typename std::allocator<T>::reference;
using const_reference = typename std::allocator<T>::const_reference;
using size_type = typename std::allocator<T>::size_type;
using difference_type = typename std::allocator<T>::difference_type;
template <typename U>
struct rebind {
typedef aligning_allocator<U> other;
};
pointer allocate(size_type n, const void* hint = 0) {
pointer result = std::allocator<T>::allocate(n, hint);
if (result) {
// Ensure the allocated memory is aligned
uintptr_t addr = reinterpret_cast<uintptr_t>(result);
uintptr_t alignment = alignof(T);
if (addr % alignment != 0) {
std::allocator<T>::deallocate(result, n);
result = std::allocator<T>::allocate(n + alignment - (addr % alignment));
if (!result) {
throw std::bad_alloc();
}
std::memset(result, 0, n * sizeof(T));
}
}
return result;
}
void deallocate(pointer p, size_type n) noexcept {
std::allocator<T>::deallocate(p, n);
}
};
int main() {
aligning_allocator<int> allocator;
std::vector<int, aligning_allocator<int>> vec;
vec.reserve(10);
for (int i = 0; i < 10; ++i) {
vec.push_back(i);
}
for (const auto& elem : vec) {
std::cout << elem << " ";
}
return 0;
}
在这个示例中,我们创建了一个名为aligning_allocator
的自定义分配器,它继承自std::allocator<T>
。我们重载了allocate
方法,以确保分配的内存是对齐的。然后,我们使用这个自定义分配器创建了一个std::vector<int>
,并向其中添加了一些元素。最后,我们遍历向量并打印其内容。