C真核引起分割断层

Question

我使用以下函数分配内存：

int qmem_alloc(unsigned int num_bytes, void ** rslt){

void** temp;
if(rslt == NULL)
    return -1;
temp = (void **)malloc(num_bytes);
if(temp == NULL)
    return -2;
else
    rslt = temp;
    return 0;
}

以及以下重新分配内存的功能：

int  qmem_allocz(unsigned num_bytes, void ** rslt){
void** temp;
void *test = (void *)malloc(10);
if(rslt == NULL)
    return -1;
temp = (void **)realloc(rslt, num_bytes);
printf("here");
if(temp == NULL)
    return -2;
else
    // free(rslt)

    return 0;
  }

以下是我的主要功能：

struct qbuf { int idx; char data[256]; };
void main(){
struct qbuf * p = NULL;
printf("%d\n",qmem_alloc(sizeof(struct qbuf), (void **)&p));
printf("%d\n",qmem_allocz(100*sizeof(struct qbuf), (void **)&p));
}

该程序可以得到内存分配，但它崩溃时，重新分配完成。以下是错误：

malloc.c:2868: mremap_chunk:断言‘((大小+偏移)& (GLRO (dl_pagesize) - 1)) == 0’失败。

为什么会发生这种情况？我怎么才能修好它？

Answer 1

您在qmem_alloc中的分配是错误的。

temp = (void **)malloc(num_bytes); //You are wrongly typecasting, don't typecast the malloc return.
rslt = temp; // This makes rslt points to object where temp is pointing

你只需要按下面的方式来做。

int qmem_alloc(unsigned int num_bytes, void ** rslt){
  if(rslt == NULL)
    return -1;

   *rslt = malloc(num_bytes);
   if(*rslt == NULL && num_bytes > 0)
      return -2;
   else
      return 0;
}

你的重新分配是错误的

temp = (void **)realloc(rslt, num_bytes); //You need to pass the object where rslt is pointing.

用于重新分配的示例代码：

int  qmem_allocz(unsigned num_bytes, void ** rslt){
   void* temp; // No pointer to pointer is needed

   void *test = (void *)malloc(10);
   if (test == NULL) return -3;

   if(rslt == NULL)
      return -1;

   temp = realloc(*rslt, num_bytes); //use *rslt to pass the address of object where rslt is pointing.

   if(temp == NULL && num_bytes > 0){
      return -2;
    }
    else{
     *rslt = temp;
      return 0;
    }
}