c - why the Push() function in my Linked List implementation shows this behaviour? -


i wrote push() function push element inside linked list. below code, wrote first :

struct node {     int data;     struct node *next; };  void push(struct node*, int);  int main(int argc, char** argv) {      struct node *list = null;      push(list, 23);      return 0; }  void push(struct node *list, int data) {     struct node *newnode;      newnode = (struct node*)malloc(sizeof(struct node));     newnode->data = data;     newnode->next = list;     list = newnode;  }

the code compiles , executes no matter how many numbers push on list, changes doesn't reflect , list remains empty.

then created function other way , used double pointer this: 

struct node {     int data;     struct node *next; };  void push(struct node**, int);  int main(int argc, char** argv) {      struct node *list = null;      push(&list, 23);      return 0;  }  void push(struct node **list, int data) {     struct node *newnode;      newnode = (struct node*)malloc(sizeof(struct node));      newnode->data = data;     newnode->next = *list;     *list = newnode; }

here code worked , pushed element on list , change reflected well.

so, why show such different behaviour, though did small test , found both 

*list

and

 list

contains same address. here did test:

struct node {     int data;     struct node *next; };   void checkaddress1(struct node *list) {     printf("address1 is: %p\n", list); }  void checkaddress2(struct node **list) {     printf("address3 is: %p\n", *list); }  int main(int argc, char** argv) {      struct node *list = null;      checkaddress1(list);     checkaddress2(&list);      return 0; }

and output was

address1 is: (nil) address3 is: (nil)

this because both pointing towards same location. means both should work. why different working.can guys elaborate? thanks

list in main() has address of first element of list. *list gives content in memory location pointed list.

in first program, passing list push() value. means list in push() merely copy of list in main() , modifications made list of push() not reflected 1 in main().

in second program, passing pointer list in main() argument push() , in push() manipulate value @ memory location pointed pointer modifying list in main().

hence changes make in push() reflected main() program 2 not program 1.


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