typedef struct Node
{
void ** pointers;
Value ** keys;
struct Node * parent;
bool is_leaf;
int num_keys;
struct Node * next;
} Node;
typedef struct ScanManager {
int keyIndex;
int totalKeys;
Node * node;
} ScanManager;
当我尝试为我的结构ScanManager分配内存时,我收到错误“Segmentation fault(core dumped)”。我在写 -
ScanManager * scanmeta = (ScanManager *) malloc(sizeof(ScanManager));
我尝试使用calloc
代替malloc
,但这不起作用。我如何为该结构分配内存,因为我想在我的代码中进一步使用它?
typedef struct Value {
DataType dt;
union v {
int intV;
char *stringV;
float floatV;
bool boolV;
} v;
} Value;
另外,我还有一个结构 -
typedef struct BT_ScanHandle {
BTreeHandle *tree;
void *mgmtData;
} BT_ScanHandle;
我在下面提到的函数中传递此结构的引用,并尝试在此处访问我的ScanManager结构 -
openTreeScan(BT_ScanHandle **handle)
{
struct ScanManager *scanmeta = malloc (sizeof (struct ScanManager));
(** handle).mgmtData = scanmeta;
/* Remaining code */
}
答案 0 :(得分:1)
我终于找到了错误。它不是在为ScanManager
分配空间。相反,我试图初始化BT_ScanHandle
结构的一些成员而不为自己分配内存空间。工作代码是 -
openTreeScan(BT_ScanHandle **handle)
{
struct ScanManager *scanmeta = malloc(sizeof(ScanManager));
*handle = malloc(sizeof(BT_ScanHandle)); //Allocating some space
(*handle)->mgmtData = scanmeta; // Initializing
/* Remaining code */
}