Reputation: 99
In the books I read that the syntax for malloc is malloc(sizeof(int)) but in one of doubly linked list program I see the following:
newnode=(struct node *)malloc(sizeof(struct node))
What is (struct node*) doing here? What is this entire code doing? btw, the code for the struct in the program is as below.
struct node
{
char line[80];
struct node *next,*prev;
};
struct node *start=NULL,*temp,*temp1,*temp2,*newnode;
Thank you
Upvotes: 3
Views: 23288
Reputation: 20794
The code is dynamically creating a pointer
to a single type of struct node
. In most versions of C, the (struct node *)
cast is not needed, and some argue that it shouldn't be used. If you remove the cast, it will be a void*
, which can be used for any type.
Therefore:
newnode = (struct node*)malloc(sizeof(struct node));
is roughly equivalent to:
newnode = malloc(sizeof(struct node));
See: Specifically, what's dangerous about casting the result of malloc?
Note 1: If you are using Visual Studio to write your C code, it will give you red underlining if you don't cast the result of malloc
. However, the code will still compile.
Note 2: Using malloc
in C++ code requires you to cast the result as shown in your example.
Upvotes: 5
Reputation: 3819
malloc returns a void pointer .
(struct node*) does a explicit conversion from void pointer to the target pointer type
Upvotes: 2
Reputation: 70939
You should pass the number of bytes that you want malloc
to allocate as argument. That is why in this code sample you use sizeof(struct node)
telling C
to allocate the number of bytes needed for a struct node variable. Casting the result like is shown in this code is bad idea by the way.
Upvotes: 2
Reputation: 11515
"malloc" returns a void-pointer. (struct node *) is type-casting the result of malloc to a "node struct pointer", which is (undoubtibly) what "newnode" is.
Upvotes: 1
Reputation:
You ran into a very bad code. C programmers never cast a result of malloc()
. Not only it is not needed but can be harmful.
Upvotes: 3