当我将一个函数指针传递给几个结构时,我得到了一个SegFault,我无法弄清楚我做错了什么。这是代码:
typedef int (*CompareFuncT)( void *, void * );
typedef void (*DestructFuncT)( void * );
struct AVL
{
void * obj;
struct AVL * parent;
struct AVL * leftChild;
struct AVL * rightChild;
};
typedef struct AVL * AVLPtr;
struct SortedList
{
AVLPtr root;
CompareFuncT comp;
DestructFuncT dest;
};
typedef struct SortedList * SortedListPtr;
SortedListPtr SLCreate(CompareFuncT cf, DestructFuncT df){
SortedListPtr slp = malloc(sizeof(struct SortedList));
if(slp == NULL){
printf("Not enough space for list\n");
return NULL;
}
slp->root = NULL;
slp->comp = cf;
slp->dest = df;
return slp;
}
AVLPtr avl_insert(AVLPtr root, AVLPtr parent, void * obj, int (*compare)( void *, void * )){
int s = 5;
int k = 6;
compare(&s, &k);
if(root == NULL){
root = malloc(sizeof(struct AVL));
if(root == NULL){
printf ("Out of memory - creating AVL node\n");
return NULL;
}
root->obj = obj;
root->parent = parent;
root->leftChild = NULL;
root->rightChild = NULL;
return root;
}
else if (compare(obj, root->obj) < 0){
root->leftChild = avl_insert(root->leftChild, root, obj, compare);
root = balance(root);
}
else if (compare(obj, root->obj) >= 0){
root->rightChild = avl_insert(root->rightChild, root, obj, compare);
root = balance(root);
}
return root;
}
int SLInsert(SortedListPtr list, void * newObj){
list->root = avl_insert(list->root, newObj, list->comp);
if(list->root == NULL)
return 0;
return 1;
}
int compareInts(void *p1, void *p2)
{
int i1 = *(int*)p1;
int i2 = *(int*)p2;
return i1 - i2;
}
void destroyBasicTypeNoAlloc(void *p) {
return;
}
int main(int argc, char **argv) {
int s = 9;
SortedListPtr list = SLCreate(compareInts, destroyBasicTypeNoAlloc);
SLInsert(list, &s);
return 0;
}
显然有更多参数通过该函数,但这是我的比较函数的传播。我在avl_insert中获得了比较的SegFault。我有一种感觉,我只是没有将指针传递到我应该的位置,但我无法找到它。
答案 0 :(得分:1)
错误是您致电malloc
:
SortedListPtr slp = malloc(sizeof(SortedListPtr));
您正在分配指针占用的字节数,这是不正确的。它应该是:
SortedListPtr slp = malloc(sizeof(struct SortedList));