C 将命令行参数组合到链接列表中。然后它被打印出来，然后释放

在字符串的链接列表中保留节点

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

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要创建并返回从“argv[1]”到“argv[argc-1]”的字符串链接列表，或者如果“argc”正在为列表分配新的ptrNode而不是追加，则返回“NULL”

每次为列表分配新的ptrNode时，上一个ptrNode都会被擦除。您需要将新的ptrNode追加到列表的末尾，并将end更新到当前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;
  }    
}

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你好，谢谢你的回答。我确实用你告诉我的方法纠正了我的错误，但是它仍然像以前一样。我已经尝试过这个代码。它起作用了。您正在返回“列表”，对吗？是的，我正在返回列表。您的代码在发布方法中也有问题。您只释放列表中的第一个节点。您需要遍历整个列表并释放每个节点。检查已编辑的post@R.. 如果回答了您的问题，请单击“接受”按钮

   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);

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

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

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

   ./argList hello there
   hello
   there

   ./argList  hello there everyone 
   hello
   there 
   everyone

    ./argList hello there 
     there

    ./argList hello there everyone
     everyone

    list = 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;
  }    
}

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);
     }
}