c - memcpy中的EXC_BAD访问权限

我试图建立一个BST并在其中插入节点。但是在创建新节点时，我不断遇到exc_bad访问错误，这可能是什么原因？这是我的代码：

struct Node *node_create(struct BSTree *bst,void *nodeKey, struct Value *nodeVal, struct     Node *rightChild, struct Node *leftChild)
{
struct Node *node = malloc(sizeof *node);
nodeKey= malloc (sizeof (bst->key_size));
nodeVal = malloc(sizeof(bst->value_size));
size_t sizeKey = sizeof(nodeKey);
memcpy(node->key, nodeKey, sizeKey);  // exc_bad access
size_t sizeVal = sizeof (nodeVal);
memcpy(node->val, nodeVal, sizeVal); // exc_bad access
node->right = rightChild;
node->left = leftChild;

return node;

}

struct Node {
void *key;
struct Value *val;
struct Node *left;
struct Node *right;
};



struct BSTree {
size_t key_size, key_alignment;
size_t value_size, value_alignment;
int (*compare_func)(void *, void *);
struct Node *root;
// ... Maybe some other stuff.
};

struct Value {
char name[10];
int id;
};

最佳答案

如果不查看Node结构，我想您想做的是：

如果节点定义为

struct Node {
    void *key;
    struct Value *val;
    struct Node *right;
    struct Node *left;
};

然后

struct Node *node_create(struct BSTree *bst,void *nodeKey, struct Value *nodeVal, struct     Node *rightChild, struct Node *leftChild)
{
struct Node *node = malloc(sizeof *node);

  node->key = malloc(bst->key_size);          /* No sizeof here */
  node->val = malloc(bst->value_size);
  memcpy(node->key, nodeKey, bst->key_size);
  memcpy(node->val, nodeVal, bst->value_size);
  node->right = rightChild;
  node->left = leftChild;

  return node;
}

由于您无需检查malloc的返回值（这是可以证明其合理性的设计选择），因此您甚至可以用这种方式编写它。

struct Node *node_create(struct BSTree *bst,void *nodeKey, struct Value *nodeVal, struct     Node *rightChild, struct Node *leftChild)
{
struct Node *node = malloc(sizeof *node);

  node->key = memcpy(malloc(bst->key_size)  , nodeKey, bst->key_size);
  node->val = memcpy(malloc(bst->value_size), nodeVal, bst->value_size);
  node->right = rightChild;
  node->left = leftChild;
  return node;
}

有些人对此风格有些畏缩，但我宁愿不要在冗余上过多地稀释我的代码。