c-为什么我的链表程序会给出SIGSEV

Question

我正在创建一个简单的链表程序来插入和查看LL的元素。当我试图插入第二个元素时，它给出了SIGSEV，但我不明白为什么？？！！

请帮我指出问题：

main.c：

#include <stdio.h>
#include <stdlib.h>
typedef struct linkedList{
    int data;
    struct linkedList *next;
}LL;
LL *start;
int main(){
    int option = 0;
    while(1){
    printf("Enter option: n");
    scanf("%d", &option);
    switch(option){
       case 1:
         addNode(start);
         break;
       case 2:
         readNodes(start);
         break;
       case 3:
         exit(0);
    }
    }
}

插入节点：

int addNode(LL *startNode){
    LL *temp, *node;
    int data = 0;
    printf("Enter data: ");
    scanf("%d", &data);
    if(startNode == NULL){/*Application only when the first element is inserted*/
        startNode = (LL*)malloc(sizeof(LL*));
        if(startNode == NULL){
          printf("Error: Memory not allocated!!n");
          exit(1);
        }
        startNode->data = data;
        startNode->next = NULL;
        start = startNode;
        return 0;
    }
    temp = startNode;
    while(temp != NULL){
        temp = temp->next;
    }
    node = (LL*)malloc(sizeof(LL*));
    node->data = data;
    node->next = NULL;
    temp->next = node;
    temp = temp->next;
    return 0;
}

Answer 1

1. sizeof占用结构的长度，对于32位体系结构，您总是传递4个字节
2. 迭代temp节点的while循环是错误的，您应该检查next节点是否为NULL