c - 复制指针时出现段错误

Question

下面的代码为什么会导致段错误？基本上在我将头指针复制到 temp 之后，头指针就消失了。

typedef struct address * paddress; // defines struct pointer

void addAddressToList(paddress head, int addr[])
{
  if (head == NULL) {
    //head->addrArray = addr; // if list is initially empty
  } else {
    paddress temp;
    temp = head;
    while (temp->right != NULL) {
      temp = temp->right;  // go to end of the list
    }
    paddress newAddress = (paddress)malloc(sizeof(paddress*));
    newAddress->intAddr = addr;
    newAddress->right = NULL;    
    newAddress->left = temp;  // connect the new address
    temp->right = newAddress;
  }
}

main() {
  paddress addressListHead;
  addressListHead = (paddress)malloc(sizeof(paddress*));
  int intAddr1[] = {1,2,3,4,5,6,7};
  char hexAddr1[] = "123456";
  int intAddr2[] = {16,14,13,12,11};
  char hexAddr2[] = "fedcb";

  addressListHead->intAddr = intAddr1;
  addressListHead->hexAddr = hexAddr1;
  addAddressToList(addressListHead, intAddr2);
}

Answer 1

paddress addressListHead;
addressListHead = (paddress)malloc(sizeof(paddress*));

似乎摆脱了编译错误，你已经将 malloc 返回到 paddress 的类型进行了类型转换。 addressListHead 是一个指针，这意味着它可以保存对象的地址，但不是指针的地址。 malloc here 语句不会创建对象。你需要改变这个-

addressListHead = (paddress)malloc(sizeof(paddress*));

到

addressListHead = (paddress)malloc(sizeof(struct address));

在 main 和 addAddressToList 函数中。

段错误:

else {
  paddress temp;
  temp = head;
  while (temp->right != NULL) {
    temp = temp->right;  // go to end of the list
}

我知道 paddress::right 是一个指针，您将它与 NULL 进行比较。但是什么是 temp::right 被初始化为。它指向一些垃圾地址，因此您不能要求它与NULL 进行比较。使其指向有效的内存位置。