'list' undeclared (first use in this function) - c

Question

I have below code.
I am getting error as 'list' undeclared (first use in this function).
Please help me
#include <stdio.h>
#include <stdlib.h>
struct list{
int data;
struct list *next;
};
typedef struct list *head;
int main()
{
struct list *start;
int i;
start = (list *) malloc(sizeof(struct list));
printf("\nEnter the data : \n");
scanf("%d", &i);
start->data = i;
start->next = NULL;
while(list->next != NULL)
{
printf("%d ", list->data);
list = list->next;
}
return 0;
}

Answer 1

You're using the type list instead of variable name start. Proper code:
while (start->next != NULL)
{
start = start->next;
// etc.
}

Answer 2

Don't cast the return type of malloc - there's no benefit of it and in this case you did it wrong!
start = (list *) malloc(sizeof(struct list));
should be
start = malloc(sizeof(struct list));
the type list * doesn't exist; you meant struct list *.
You can make it even more safe by writing
start = malloc(sizeof(*start));
this way you automatically malloc enough bytes for the (pointer) type of start, which is useful when you later change the type of start - the malloc call doesn't change a bit.

Answer 3

The
start = (list *) malloc(sizeof(struct list));
includes an unnecessary typecast. Just do
start = malloc(sizeof(struct list));
However, your code has more trouble than this. I can answer your question best by asking a question of my own: in your mind, is list a type or an object?
If you answer this question, one suspects that you can fix your code. Good luck.

Answer 4

#include <stdio.h>
#include <stdlib.h>
typedef struct list{
int data;
struct list *next;
} list;
typedef struct list *head;
int main()
{
struct list *start;
int i;
start = (list *) malloc(sizeof(struct list));
printf("\nEnter the data : \n");
scanf("%d", &i);
start->data = i;
start->next = NULL;
while(start->next != NULL)
{
start = start->next;
}
return 0;
}
you can define the type (list *)

Answer 5

Your variable is named « start » and you called it « list ».

Answer 6

start = (struct list *) malloc ...
You've missed struct in casting.
Which is not neccessary at all as Anthales pointed out.
start = malloc ...

Answer 7

One problem you had is you were re-using struct in declaring your struct pointer after you had created a typedef, struct list *start;. Also the struct and typedef cannot have the same name. You get this:
cc -Wall test.c -o test
test.c: In function ‘main’:
test.c:13: error: ‘list_t’ undeclared (first use in this function)
test.c:13: error: (Each undeclared identifier is reported only once
test.c:13: error: for each function it appears in.)
test.c:13: error: ‘start’ undeclared (first use in this function)
test.c:13: error: ‘cur’ undeclared (first use in this function)
test.c:13: warning: left-hand operand of comma expression has no effect
test.c:16: error: expected expression before ‘)’ token
You can choose to use struct list everywhere and skip making using typedef. Use of typedef simplifies how your code reads as noted here: http://en.wikipedia.org/wiki/Struct_%28C_programming_language%29#typedef
I've rewritten what you have just so I could compile it and understand it a little better, and so I could put some data into one node. I remember the whole struct typedef concept taking a little time to sink in, when I was learning C. So, don't give up.
#include <stdio.h>
#include <stdlib.h>
struct list {
int data;
struct list *next;
};
typedef struct list list_t;
int main()
{
list_t *start, *cur;
int i;
start = (list_t *) malloc(sizeof(list_t));
if (NULL != start)
{
cur = start; /* Preserve list head, and assign to cur for list trarversal. */
printf("\nEnter the data : ");
scanf("%d", &i);
cur->data = i;
cur->next = NULL;
cur = start;
while(cur != NULL)
{
printf("%d ", cur->data);
cur = cur->next;
}
}
else
{
printf("Malloc failed. Program ending.");
}
return 0;
}