how to use a pointer inside struct in C

Question

I have a two struct, one is linked list.

typedef struct Mark{
       int  people;
       Node *nodeyy;
}Mark

typedef struct Node{
       struct node next;
       int value;
}Node

if i allocated memory for a node, let say

 Node *node1=malloc( sizeof(struct Node));

And I also allocated memory for a bookmark, let say

 Mark *mark1=malloc( sizeof(struct Mark));

I want to make the pointer nodeyy in the mark1 points to the same thing as node1, how can i do that?

I think that

 mark1->nodeyy=node1;

is definitely wrong.

Now that you have edited the code, why do you think mark1->nodeyy = node1 is wrong? — Art Taylor
– Art Taylor, Commented Oct 20, 2012 at 7:25

Rajesh Pantula · Accepted Answer · 2012-10-20 07:17:12Z

1

change the int* in struct Mark to Node*

typedef struct Mark{
       int  people;
       Node *nodeyy;
}Mark

then you can do

mark -> nodeyy =  (Node *) malloc(sizeof(Node))

answered Oct 20, 2012 at 7:17

Rajesh Pantula

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6 Comments

runcode Over a year ago

i though mark-> nodeyy just a reference to a node, why we need to allocate memory for a reference? What is the reason we can not just say mark->nodey=node1; ?

Rajesh Pantula Over a year ago

ya but mark->nodeyy is an int* in your code and it cannot hold the address of struct Node

runcode Over a year ago

sorry, typo , i mean Node *nodeyy;

runcode Over a year ago

we can say mark1->people=10, can we say mark1->nodeyy= node1; ? My question: is that the malloc steps necessary?

Rajesh Pantula Over a year ago

because node1 is just a type(just like primitive types int,float). you need to create variables of that type and make mark->nodeyy point to that variable.

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Aniket Inge · Accepted Answer · 2012-10-20 07:27:22Z

1

its correct now:

You will have to initialize the pointer or point it to an existing variable that you know won't go out-of-scope. BUT since node1 is dynamically allocated, you're just assigning one pointer to another, this creates a sort of reference to the newly allocated memory pointed by node1.

mark1->nodeyy = node1;

After this statement, mark1->nodeyy and node1 point to the memory location returned by the malloc(sizeof(Node)).

edited Oct 20, 2012 at 7:27

answered Oct 20, 2012 at 7:14

Aniket Inge

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1 Comment

runcode Over a year ago

sorry, typo , i mean Node *nodeyy;

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