有可能在你的code其他错误，但一个大问题是，你正试图设置头节点插入，但只影响传入的指针的本地副本，所以它在主叫方没有任何影响：

There may be other errors in your code, but one big issue is that you are attempting to set a head node in insert, but that only affects a local copy of the pointer passed in, so it has no effect in the caller side:

void  insert(struct node *head,int data){
  ....
  head = malloc(sizeof(struct node)); // head is local, caller won't see this

您还需要确保当你通过这不是一个节点 NULL ，你居然attatch新节点的头部。
您可以通过传递一个指针的指针，或返回集指针解决的第一个问题。例如，

You also need to ensure that when you pass a node that is not NULL, you actually attatch the new node to the head. You can fix the first problem by passing a pointer to a pointer, or by returning the set pointer. For example,

void insert(struct node **head, int data) {
  if(*head == NULL) {
    // create the head node
    ...
    *head = malloc(sizeof(struct node)); 
    ....
  else {
    // create a new node and attach it to the head
    struct node* tmp = malloc(sizeof(struct node));
    ....
    (*head)->next = tmp;
  }
}

然后，在主，你需要一个指针传递给头指针，即使用地址运算符＆安培; ：

Then, in main, you need to pass a pointer to the head pointer, i.e. use the address-of operator &:

struct node *head = NULL;
insert(&head, 5);

注意的部分问题是该函数试图做太多。这就是所谓的插入，但它试图创建一个新的节点，如果传入的指针 NULL 。这将是更好的分离这些职责：

Note part of the problem is that the function is trying to do too much. It is called insert, but it attempts to create a new node if the pointer passed in is NULL. It would be better to separate these responsibilities:

// allocate a node and set its data field
struct node* create_node(int  data)
{
  struct node* n = malloc(sizeof(struct node));
  n->next = NULL;
  n->data = data;
  return n;
}

// create a node and add to end node.
// return the new end node.
// end has to point to a valid node object.
struct node* append_node(struct node* tail, int node_data)
{
  struct node* new_tail = create_node(node_data);
  tail->next = new_tail;
  return new_tail;
}