动态内存分配，地址对齐

转载的一个小程序，学习学习。

align_malloc函数返回的align地址可按任意字节对齐。分配size+align_ment的空间，多分配alignment空间主要是用来调整指针的地址。调整好以后将mem_ptr指针与tmp指针之间的偏移量保存在*(memptr-1)的内存地址处，回收内存时用来找到真正的内存块地址即tmp指针。

 i386中堆的分配是向上增长的，而栈则上向下增长。

 

这里有点不明白，为什么要把这个偏移量保存在*(memptr-1)中，保存在*(memptr-2)或者一个全局指针中应该都可以吧？？

 我们可以选择保存偏移量，也可以选择保存指向tmp指针的指针。

 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void * align_malloc(unsigned int size, unsigned int alignment)
{
    unsigned char * mem_ptr;
    unsigned char * tmp;

    if(!alignment) alignment=4; //至少按4对齐
    /* Allocate the required size memory + alignment so we
    * can realign the data if necessary */
    if ((tmp = (unsigned char *) malloc(size + alignment)) != NULL){

        /* Align the tmp pointer */
        mem_ptr = (unsigned char *) ((unsigned int) (tmp + alignment - 1) & (~(unsigned int) (alignment - 1)));

        /////////mem_ptr= (unsigned char*)tmp + (alignment - (unsigned long)(tmp) % alignment);也可以按这种方法来调整指针

        /* Special case where malloc have already satisfied the alignment
        * We must add alignment to mem_ptr because we must store
        * (mem_ptr - tmp) in *(mem_ptr-1)
        * If we do not add alignment to mem_ptr then *(mem_ptr-1) points
        * to a forbidden memory space */
        if (mem_ptr == tmp)
            mem_ptr += alignment;

        /* (mem_ptr - tmp) is stored in *(mem_ptr-1) so we are able to retrieve
        * the real malloc block allocated and free it in xvid_free
  mem_ptr-1这个内存地址用来存储调整后的mem_ptr与真正的内存分配块之间的偏移量*/
        *(mem_ptr - 1) = (unsigned char) (mem_ptr - tmp);
        //PRT("Alloc mem addr: 0x%08x, size:% 8d, file:%s <line:%d>, ", tmp, size, file, line);
        /**//* Return the aligned pointer */
  printf("memptr address is %p\n",mem_ptr);
  printf("tmp address is %p\n",tmp);
  printf("offset is %d\n",*(mem_ptr -1));
        return ((void *)mem_ptr);
    }
    return(NULL);
}

/**//*****************************************************************************
* align_free
*
* Free a previously 'xvid_malloc' allocated block. Does not free NULL
* references.
*
* Returned value : None.
*
****************************************************************************/

void align_free(void *mem_ptr)
{
    unsigned char *ptr;
    if (mem_ptr == NULL)
        return;

    /**//* Aligned pointer */
    ptr = (unsigned char *)mem_ptr;

    /* *(ptr - 1) holds the offset to the real allocated block
    * we sub that offset os we free the real pointer
 减去这个偏移值，找到真正的内存分配块地址*/
    ptr -= *(ptr - 1);

    /**//* Free the memory */
    free(ptr);
}

int main()
{
 int *ptr;
 ptr = (int *)align_malloc(256,24);
 printf("address is %p\n",ptr);
 printf("value is %d \n",(unsigned int)ptr % 24);
 align_free(ptr);
 return 0;
}

 

程序的打印来看，返回的地址的确是按24字节对齐的。