2016-08-30 24 views
0

我在C中用-O3编译,在Linux Fedora x64上用作共享库。我之后使用了一些malloc。似乎当我在这个家庭工作与wmemcpy和其他功能在一个内存的变化使其他更改... 我可能会违反内存大小?或者问题出在我的实现中?有人可以解释我的记忆池中有什么问题吗?

在此先感谢!

char* ur_free_available(struct memory_pool* pool, uint64_t sz) 
{ 
    uint64_t size = sz + sizeof(struct _MyPage); 
    if(pool->free_pos > 0) 
    { 
     if(pool->free_blocks[pool->free_pos - 1].pNext - pool->free_blocks[0].pNext > size) 
     { 
      uint64_t i, j; 
      char *addr; 
      struct _MyPage mpg; 
      for(i = 1; i < pool->free_pos; ++i) 
      { 
       if(pool->free_blocks[i].pNext - pool->free_blocks[0].pNext > size) 
       { 
        mpg.pPrev = NULL; 
        mpg.pNext = pool->free_blocks[i-1].pNext; 
        break; 
       } 

      } 
      pool->free_blocks[0].pPrev = NULL; 
      for(j = i; j < pool->free_pos; ++j) 
      {     
        if(j > 0) 
         pool->free_blocks[j-i].pPrev = pool->free_blocks[j-1].pPrev; 
        pool->free_blocks[j-i].pNext = pool->free_blocks[j].pNext;          
      } 
      pool->free_pos -= i; 
      pool->free_pos++; 
      memcpy(addr,(char*)&mpg, sizeof(mpg)); 
      return &addr[0] + sizeof(struct _MyPage); 
     } 
    } 
    return NULL; 
} 


char* ur_mem_alloc(struct memory_pool* pool, uint64_t sz) 
{ 
    char *addr; 
    struct _MyPage mpg; 
    uint64_t size = sz + sizeof(mpg); 
    uint64_t j = 0, c, op; 
    if(pool->init_pos_is != 7) 
    { 
     //bzero(pool, sizeof(pool));   
     pool->init_pos_is = 7; 
     pool->free_pos = 0; 
     pool->hp_position = 0; 
     pool->st_position = 0; 
     mpg.pPrev = NULL; 
     for(c = 0; c < HP_CNT; c++) 
      pool->_extramem[c] = NULL; 
     pool->free_blocks[0].pNext = NULL; 
     //pool->_bytes = malloc(ST_MYPAGE_NO*ST_MYPAGE_SIZE); 
     //pthread_mutex_lock(&llk1); 
     _mpool = pool; 
     //pthread_mutex_unlock(&llk1); 
     atexit(equit); 
    } 

    if((addr = ur_free_available(pool, size)) != NULL)  
    { 
     return &addr[0]; 
    } 
    op = pool->hp_position; 
    if(size + (ST_MYPAGE_SIZE * pool->st_position) > ST_MYPAGE_NO * ST_MYPAGE_SIZE) 
    { 
     uint64_t i;  
     for(i = 0; i < HP_CNT; ++i) 
     { 
      if(size < (pow(8, i)-pool->hp_position)*HP_MYPAGE_SIZE)    
      { 
       j = i;    


       if(pool->_extramem[i] == NULL) 
       {  
        pool->hp_position = 0; 
        pthread_mutex_lock(&llk2); 
        pool->_extramem[i] = (char*)malloc((uint64_t)pow(8, i) * HP_MYPAGE_SIZE); 
        pthread_mutex_unlock(&llk2); 
       } 
       break;   
      } 

     } 

     addr = &pool->_extramem[j][pool->hp_position*HP_MYPAGE_SIZE]; 
     mpg.pPrev = (struct _MyPage*)&pool->_extramem[j][op*HP_MYPAGE_SIZE]; 
     //printf("j %u %u %u\n", j, (uint64_t)size, (uint64_t)(pow(8, i-1))*HP_MYPAGE_SIZE); 
     pool->hp_position += floor(size/HP_MYPAGE_SIZE) + 1;   
     mpg.pNext = (struct _MyPage*)&pool->_extramem[j][pool->hp_position*HP_MYPAGE_SIZE]; 

     memcpy(addr,(char*)&mpg, sizeof(mpg)); 
     return &addr[0] + sizeof(struct _MyPage); 
     // 
    } 
    else 
    { 
     if(pool->st_position != 0) 
     { 
      mpg.pPrev = (struct _MyPage*)&pool->_extramem[j][(pool->hp_position)]; 
     } 
     mpg.pNext = NULL; 
     pool->st_position += floor(size/ST_MYPAGE_SIZE); 
     addr = &pool->_bytes[(uint64_t)(pool->st_position-floor(size/ST_MYPAGE_SIZE)) * ST_MYPAGE_SIZE]; 
     memcpy(addr,(char*)&mpg, sizeof(mpg)); 
     return &addr[0] + sizeof(struct _MyPage); 
    } 

} 
void ur_mem_free(struct memory_pool* pool, char *addr) 
{ 
    if(addr == NULL) return; 

    pool->free_blocks[pool->free_pos].pPrev = pool->free_blocks[pool->free_pos].pNext; 
    pool->free_blocks[pool->free_pos].pNext = (struct _MyPage*)(addr - sizeof(struct _MyPage)); 
    pool->free_pos++; 
} 
+0

你正在努力实现你的实现? – Serge

+0

不会调用malloc很多...(并且在未满时更快的内存使用堆栈) –

+0

malloc的性能已经是瓶颈了吗?什么让你认为你可以自由地将malloc返回的指针解释为指向单个连续的内存块的指针,这些指针都可以使用(我的意思是你在ur_free_available函数中使用的技巧)?你有没有在glibc的源代码中检查malloc的实现? – Serge

回答

0

Breafly寻找到功能ur_free_available,变量addr没有初始化,但你的memcpy把它损坏了一个随机的内存区域。然后您尝试使用相同的非初始化指针返回某些内容。

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