std::align

一、其实就是实现了align to功能

cppreference std::align

Example

#include <iostream>
#include <memory>

template<std::size_t N>
struct MyAllocator
{
    char data[N]; // memory pool
    void *p;
    std::size_t sz; // memory pool的剩余空间的大小
    MyAllocator() :
                    p(data), sz(N)
    {
    }
    template<typename T>
    T* aligned_alloc(std::size_t a = alignof(T))
    {
        if (std::align(a, sizeof(T), p, sz))
        {
            T *result = reinterpret_cast<T*>(p); // p指向分配的内存起始地址
            p = (char*) p + sizeof(T); // 更新p
            sz -= sizeof(T); // 更新sz
            return result;
        }
        return nullptr;
    }
};

int main()
{
    MyAllocator<64> a;

    // allocate a char
    char *p1 = a.aligned_alloc<char>();
    if (p1)
        *p1 = 'a';
    std::cout << "allocated a char at " << (void*) p1 << '\n';

    // allocate an int
    int *p2 = a.aligned_alloc<int>();
    if (p2)
        *p2 = 1;
    std::cout << "allocated an int at " << (void*) p2 << '\n';

    // allocate an int, aligned at 32-byte boundary
    int *p3 = a.aligned_alloc<int>(32);
    if (p3)
        *p3 = 2;
    std::cout << "allocated an int at " << (void*) p3 << " (32 byte alignment)\n";
}
// g++ test.cpp --std=c++11 -pedantic -Wall -Wextra

在我的环境中的运行结果如下:

allocated a char at 0x7fffd9d98800
allocated an int at 0x7fffd9d98804
allocated an int at 0x7fffd9d98820 (32 byte alignment)