c - 将命令行参数组装成链接列表。然后打印出来，free()d

Question

在字符串链表中保存一个节点。

struct  Node
{
   char*  namePtr_;
   struct Node* nextPtr_;
};

---

struct Node*  makeList(int  argc,
             char* argv[])
   {
   struct Node* list = NULL;
   struct Node* end = NULL;
   int  i;

---

创建并返回从 'argv[1]' 到 'argv[argc-1]' 的字符串链表，或者如果 'argc' <= 1 则返回 'NULL'。

   for  (i = 1;  i < argc;  i++)
   {
     struct Node * ptrNode;
     ptrNode = (struct Node*)malloc(sizeof(*list));
     ptrNode -> namePtr_ = (char *) malloc(strlen(argv[i])+1);
     strcpy(ptrNode -> namePtr_, argv[i]);
     ptrNode -> nextPtr_ = NULL;

     list = ptrNode;  //I think my problem is here
   }  
   }
  return(list);

打印“list”中的“namePtr_”值。

    void print (const struct Node* list)
    {
      const struct Node* run;
      run = list;
      while(run != NULL){
          printf("%s\n", run->namePtr_);
          run = run -> nextPtr_;  
     }
     }

---

释放列表的 nextPtr 和 namePtr，以及所有 nextPtr_ 后继者。

  void release (struct Node*    list)
  {
     struct Node* ans = list;
     free(ans);  
  }

---

创建、打印和 free() 链表。

        int main(int argc, char*  argv[])
       {     
         struct Node*   list;
         list   = makeList(argc,argv);
         print(list);
         release(list);  
         return(EXIT_SUCCESS);

它应该输出:

   ./argList hello there
   hello
   there

   ./argList  hello there everyone 
   hello
   there 
   everyone

---

但是我的正在输出:

    ./argList hello there 
     there

    ./argList hello there everyone
     everyone

Answer 1

您正在将新的 ptrNode 分配给列表而不是追加

    list = ptrNode;

每次您将新的 ptrNode 分配给列表并且之前的 ptrNode 被删除。您需要将新的 ptrNode 附加到列表的末尾并将 end 更新为当前的 ptrNode

for (i = 1; i < argc; i++) { 
  struct Node * ptrNode;

  ptrNode = (struct Node*)malloc(sizeof(*list));

  ptrNode -> namePtr_ = (char *) malloc(strlen(argv[i])+1);

  strcpy(ptrNode -> namePtr_, argv[i]);

  ptrNode -> nextPtr_ = NULL;

  if(list==NULL){
    list = ptrNode; 
    end = ptrNode;
  }
  else {
   end->nextPtr_= ptrNode;
   end = ptrNode;
  }    
}

对于释放列表，你的代码只释放列表中的第一个节点，你需要遍历整个列表并释放每个节点

void release (struct Node*    list)
  {
     struct Node* run;
      run = list;
      while(run != NULL){
          printf("%s\n", run->namePtr_);
          struct Node* temp = run;
          run = run -> nextPtr_;  
          free(temp);
     }
}