由网友(对的时间对的人)分享简介：你怎么分配多数民众赞成对齐到特定边界C（例如，高速缓存行边界）的内存？我正在寻找的malloc /免费的像实施，理想将尽可能实现可移植性---至少在32位和64位架构。How do you allocate memory that's aligned to a specific boundary in C (e.g....

你怎么分配多数民众赞成对齐到特定边界C（例如，高速缓存行边界）的内存？我正在寻找的malloc /免费的像实施，理想将尽可能实现可移植性---至少在32位和64位架构。

How do you allocate memory that's aligned to a specific boundary in C (e.g., cache line boundary)? I'm looking for malloc/free like implementation that ideally would be as portable as possible --- at least between 32 and 64 bit architectures.

编辑补充：换句话说，我在寻找的东西会表现得像是的 memalign可功能，可以使用免费的释放。

Edit to add: In other words, I'm looking for something that would behave like (the now obsolete?) memalign function, which can be freed using free.

推荐答案

下面是一个解决方案，它封装在调用malloc的，分配的调整目的更大的缓冲区，只为一对准缓冲区之前存储原始分配的地址后来打电话来释放。

Here is a solution, which encapsulates the call to malloc, allocates a bigger buffer for alignment purpose, and stores the original allocated address just before the aligned buffer for a later call to free.

// cache line
#define ALIGN 64

void *aligned_malloc(int size) {
    void *mem = malloc(size+ALIGN+sizeof(void*));
    void **ptr = (void**)((long)(mem+ALIGN+sizeof(void*)) & ~(ALIGN-1));
    ptr[-1] = mem;
    return ptr;
}

void aligned_free(void *ptr) {
    free(((void**)ptr)[-1]);
}