find function in linked list in c programming - c

Question

I wanna write a program about linked list that gets a number and if the number was equal to the one of the nodes' data , gives the number of that node .
like the datas in 3 nodes are
123
56
78
and it gets a number like 56 and its equal to the second node's data so the output should be 2.
#include<stdio.h>
#include<stdlib.h>
struct node{
int data;
node *next;
};
node* cnode();
void find(node *first,int i,int d);
int main()
{
node *first,*last;
int n,i,x,d;
printf(" How many nodes ?\t");
scanf("%d",&x);
for(i=1;i<=x;i++){
last=cnode();
last->next=first;
first=last;
}
printf("\n enter a particular data:\t");
scanf("%d",&d);
printf("\n number of the particular node:\t");
find(first,i,d);
}
void find(node *first,int i,int d){
int count=0;
while (first != NULL)
{
if (first->data == d)
count++;
first = first->next;
}
if(count != 0){
printf("%d",count);
}
if(count == 0){
printf("\n NOT FOUND ! ");
}
}

Answer 1

According to the C Standard the function main without parameters shall be declared like
int main( void )
Secondly this declaration of the structure
struct node{
int data;
node *next;
};
is not a valid C declaration.
You should write
struct node{
int data;
struct node *next;
};
typedef struct node node;
or
typedef struct node{
int data;
struct node *next;
} node;
You have to initialize at least node first with NULL. Otherwise the program will have undefined behavior.
node *first = NULL,*last;
The parameter i is not used within the function find. So it may be removed.
void find(node *first, int d);
The function definition can look at least like
void find( node *first, int d )
{
int count = 0;
while ( first != NULL && first->data != d )
{
first = first->next;
++count;
}
if ( first != NULL )
{
printf("%d",count);
}
else
{
printf("\n NOT FOUND ! ");
}
}