Why pointer variable hold new memory at each function call

Why pointer variable (new_node) hold new memory at each function call in my code ?

Code:

---

  #include<stdio.h>
#include<stdlib.h>

struct node
{
    struct node *next_node;
};


struct node *addNodeToFront( struct node *list)
{
    struct node *new_node = malloc (sizeof(*new_node));
    printf("new_node = %p\n", (void *)new_node);
    if ( new_node != NULL)
    {
      //  new_node -> next_node = list;
       
    }
    return  new_node;
}


 
int main()
{
 
    struct node *list = NULL;
    list = addNodeToFront(list);
    list = addNodeToFront(list);
    list = addNodeToFront(list);
    return 0; 
}

---

new_node = 00021230
new_node = 00021260
new_node = 00021270

Because you are adding a new node to the front of the list each time.

Edit: In your case you are only allocation space for a new node each time.
I suggest reading the assignment.

Tim S.

Quote:

---

Originally Posted by stahta01

Because you are adding a new node to the front of the list each time.

Edit: In your case you are only allocation space for a new node each time.
I suggest reading the assignment.

Tim S.

---

I don't understand clearly why new memory allocate each time

malloc return pointer in function

Code:

---

#include<stdio.h>
#include<stdlib.h>

struct node
{
    struct node *next_node;
};


struct node *addNodeToFront( struct node *list)
{
    struct node *new_node = malloc (sizeof(*new_node));
    printf("new_node = %p\n", (void *)new_node);
    if ( new_node != NULL)
    {
      new_node -> next_node = list;
      printf("new_node -> next_node = %p\n", (void *) new_node -> next_node);   
    }
    return  new_node;
}


 
int main()
{
 
    struct node *list = NULL;
    printf("list = %p\n", (void *)list);
    list = addNodeToFront(list);
    printf("list = %p\n", (void *)list);
    list = addNodeToFront(list);
    list = addNodeToFront(list);
    return 0; 
}

---

list = 00000000
new_node = 00B61230
new_node -> next_node = 00000000
list = 00B61230
new_node = 00B61260
new_node -> next_node = 00B61230
new_node = 00B61270
new_node -> next_node = 00B61260

> I don't understand clearly why new memory allocate each time
Because that's what malloc does, returns a pointer to some new memory.

If you want to store a list of 10 items, you need to allocate space for 10 items somehow.

In this case, this means creating a list of 10 nodes, which will entail calling malloc (via addNodeToFront) 10 times.

Quote:

---

Originally Posted by Salem

> I don't understand clearly why new memory allocate each time
Because that's what malloc does, returns a pointer to some new memory.

If you want to store a list of 10 items, you need to allocate space for 10 items somehow.

In this case, this means creating a list of 10 nodes, which will entail calling malloc (via addNodeToFront) 10 times.

---

Code:

---

#include<stdio.h>
#include<stdlib.h>


struct node
{
    int data;
    struct node *next_node;
};


struct node *addNodeToFront( struct node *list, int value)
{
    struct node *new_node = malloc (sizeof(*new_node));


    if ( new_node != NULL)
    {
      new_node -> data = value;   
      new_node -> next_node = list;
     
    }
    return  new_node;
}


void show(struct node *c)
{
    if (c == NULL)
        return;
    show(c -> next_node);
    printf("%d\n", c-> data);
}
 
int main()
{
 
    struct node *list = NULL;
    list = addNodeToFront(list, 1);


    show(list);
    return 0; 
}

---

I know program has thses variables
list ( pointer variable type struct )
value ( variable type int)
new_node ( pointer variable type struct )
c( pointer variable type struct )

Are these variable in code ?


new_node -> data
new_node -> next_node

I think they are variables and occupy memory. They will store the data of node at function call