C中的简单结构程序

Question

I'm new to programming so please bear with me. Can anyone please explain what the last line of code is doing and what this entire block of code trying to do. I know the first block is creating a structure called node and adding two pointers *next and *prev to it. btw this is a part of a linked list program

struct node
 {
  char line[80];
  struct node *next,*prev;
 };

struct node *start=NULL,*temp,*temp1,*temp2,*newnode;

Thank you in advance.

Answer 1

struct node *start=NULL,*temp,*temp1,*temp2,*newnode;

can be as

struct node *start=NULL;
struct node  *temp;
struct node *temp1;
struct node *temp2;
struct node *newnode;

Now is it easy to understand..?

Answer 2

The last line is creating 5 pointers of type struct node namely start which is pointing to NULL , temp , temp1 , temp2 and newnode .

The whole block of code is actually creating a struct called node which contains an array of 80 characters, followed by pointers to next and previous. Hence it is creating a structure for a doubly linked list.

Answer 3

The last line is creating 5 pointers of type struct node. Here start is initilaized to Null.

Answer 4

The last line of code defines 5 variables, all of which are pointers to struct node . One of them, start , is initialized to NULL; the others are also initialized to NULL if the variables are outside any function, but are uninitialized if they are inside a function. Uninitialized pointers should be initialized before they are used. Null pointers should not be dereferenced; madness lies at the end of a null pointer (or, more usually, a core dump).