Отправлено: 22.11.14 07:15. Заголовок: structure variables c programming

I want to modify this program to use a linked list o PayRecord structure variables instead of an array of those structures:

#include <stdio.h>

#define NUMRECS 5

struct PayRecord /* construct a global structure type */
{
int id;
char name[20];
double rate;
};

int main()
{
int i;
struct PayRecord employee[NUMRECS] = {{32479, "Abrams, B.", 6.72},
{33623, "Bohm, P.", 7.54},
{34145, "Donaldson, S.", 5.56},
{35987, "Ernst, T.", 5.43},
{36203, "Gwodz, K.", 8.72}
};
struct PayRecord *recptr = employee; /* Pointer to array elements */
int idnum; /* User entered employee id value */
double rate; /* User entered pay rate */
int found = 0; /* Flag when employee record found */

for (i = 0; i < NUMRECS; i++, recptr++)
printf("%d %-20s %4.2f\n",
recptr->id,recptr->name,recptr->rate);

/* Get employee id and new rate */
printf("\nEnter employee number followed by new pay rate: ");
scanf("%d %lf", &idnum, &rate);

/* Search for employee id in array */
for (i = 0, recptr = employee; i < NUMRECS && !found; i++)
if (recptr->id == idnum)
found = 1; /* Set flag when found */
else recptr++; /* Otherwise increment pointer */

if (!found) /* No record found */
printf("\nNo such employee found!\n\n");
else /* Found the employee */
recptr->rate = rate; /* Update the rate */

/* Output the array again */
for (i = 0, recptr = employee; i < NUMRECS; i++, recptr++)
printf("%d %-20s %4.2f\n",
recptr->id,recptr->name,recptr->rate);
return 0;
}


Kind of like this:

#include <stdio.h>

int main()
{
struct TeleType /* Structure with link pointer */
{
char name[30];
char phoneNum[15];
struct TeleType *nextaddr;
};

/* A series of structure variables, or records, are declared and */
/* initialized, except for the nextaddr field: */
struct TeleType t1 = {"Acme, Sam","(555) 898 2392"};
struct TeleType t2 = {"Dolan, Edith","(555) 682 3104"};
struct TeleType t3 = {"Lanfrank, John","(555) 718 4581"};

/* Now the nexaddr field in each record is updated so that each record */
/* has a pointer to the next record. The last record has NULL in the */
/* nextaddr field to close off the list. */
t1.nextaddr = &t2; /* store t2's address in t1.nextaddr */
t2.nextaddr = &t3; /* store t3's address in t2.nextaddr */
t3.nextaddr = NULL; /* store the NULL address in t3.nextaddr */

struct TeleType *outptr = &t1; /* Point into list */

while (outptr != NULL) /* display till end of linked list */
{
printf("%-30s %-20s\n", outptr->name, outptr->phoneNum);
outptr = outptr->nextaddr; /* get next address */
}

return 0;
}


Any help would be great :)

 Отправлено: 22.11.14 08:59. Заголовок: If you need the same..

If you need the same functionality as with struct TeleType then simply add one more member to your original structure that will be a pointer to a next element in the list.

Or you could define one more structure like

 
 struct Node 
 { 
     struct Node *next; 
     struct PayRecord record; 
 }; 
 

As for me then it is a bad idea to combine local variables in a list because if you need to add a new element in the list you need to declare a new local variable with unique name.

I think that you need a list where each element of which is allocated dynamically.