Segmentation fault when trying to print out a struct - c

Question

I'm trying to make a code that will allow the user to input any amount of entries he wants, and then print them out (and other functions, still have to get to that). But when I try to launch the code it allows me to input the entries, but when I want to print them out, it doesn't register current.name or current.telNo (only prints out 1: has tel. No. ) and a segmentation error follows after it. Any idea how I can fix that.
#include <stdio.h>
#include <stdlib.h>
int listSize;
int counter = 0;
struct Entry {
char name[20];
char telNo[9];
struct Entry *next;
} current;
int main()
{
struct Entry *linkedList = (struct Entry * )malloc(sizeof(struct Entry));
struct Entry *current = linkedList;
first(current);
print(current);
return 0;
}
void first(struct Entry linkedList)
{
int i;
printf("enter list size: ");
scanf("%d", &listSize);
printf("Now enter entries one by one: \n");
for (i = 0; i < listSize; i++) {
counter++;
printf("Name: ");
scanf("%s", linkedList.name);
printf("Telephone: ");
scanf("%s", linkedList.telNo);
if (i != listSize -1) {
linkedList.next = (struct Entry *)malloc(sizeof(struct Entry));
linkedList = *linkedList.next;
} else {
linkedList.next = NULL;
}
}
}
void print(struct Entry linkedList)
{
int nr = 1;
printf("\nTelephone book is:\n");
while (current.name != NULL) {
printf("%d: %s has tel. No.\t%s\n", nr, current.name, current.telNo);
current = *current.next;
nr++;
}
}

Answer 1

Instead of . you should have used -> and in your print() you were traversing current instead of linkedList which was causing the issue. Also your functions definitions should come before its usage. Please check the below snippet, i have made the corresponding changes.
#include <stdio.h>
#include <stdlib.h>
int listSize;
int counter = 0;
struct Entry {
char name[20];
char telNo[9];
struct Entry *next;
} current;
void print(struct Entry *linkedList)
{
int nr = 1;
printf("\nTelephone book is:\n");
while (linkedList->name != NULL) {
printf("%d: %s has tel. No.\t%s\n", nr, linkedList->name, linkedList->telNo);
linkedList = linkedList->next;
nr++;
}
}
void first(struct Entry *linkedList)
{
int i;
printf("enter list size: ");
scanf("%d", &listSize);
printf("Now enter entries one by one: \n");
for (i = 0; i < listSize; i++) {
counter++;
printf("Name: ");
scanf("%s", linkedList->name);
printf("Telephone: ");
scanf("%s", linkedList->telNo);
if (i != listSize -1) {
linkedList->next = (struct Entry *)malloc(sizeof(struct Entry));
linkedList = linkedList->next;
} else {
linkedList->next = NULL;
}
}
}
int main()
{
struct Entry *linkedList = (struct Entry * )malloc(sizeof(struct Entry));
struct Entry *current = linkedList;
first(current);
print(current);
return 0;
}

Answer 2

while (current.name != NULL)
should be
while (current != NULL)
otherwise current = *current.next will become NULL and crash on the while condition

Answer 3

current.name is a pointer to reserved 20-byte area, which is always non-NULL.
I would initialize *next with NULL every time I allocate it (there are few ways to do that, easiest: assign NULL to it right after malloc). And do the check like that: if (current.next != NULL).
You also can check for current.name[0] != 0, but the first option is cleaner.

Answer 4

There are a number of errors in your code but, essentially two main 'points of confusion':
First, you seem to be confusing a structure (struct Entry) with a pointer-to-structure (as in, for example, struct Entry *next;).
Second, you have two different variables called current - one defined globally (which is the only one that is 'visible' to your print function), and another one defined locally inside main (this one will 'hide' the former).
Here is a corrected version of your code, with triple-slash (///) comments wherever I've made changes. Feel free to ask for further clarification and/or explanation.
#include <stdio.h>
#include <stdlib.h>
int listSize;
int counter = 0;
struct Entry {
char name[20];
char telNo[9];
struct Entry* next;
} *current; /// Here, we define a GLOBAL variable, "current" that is a POINTER to the struct
void first(struct Entry* linkedList); /// Put "Function Prototypes* here, so that "main" knows what they are
void print(struct Entry* linkedList); /// You need to pass POINTERS to structures, not the actual structures
int main()
{
struct Entry* linkedList = (struct Entry*)malloc(sizeof(struct Entry));
// struct Entry* current = linkedList; /// This declaration 'hides' the global variable
current = linkedList; /// Here, we (properly) assign the global pointer's value
first(current);
print(current);
return 0;
}
void first(struct Entry* linkedList) /// Pointer (see above)
{
int i;
printf("enter list size: ");
scanf("%d", &listSize);
printf("Now enter entries one by one: \n");
for (i = 0; i < listSize; i++) {
counter++;
printf("Name: ");
scanf("%s", linkedList->name); /// Use "->" instead of "." to get a pointer's member (and elsewhere)
printf("Telephone: ");
scanf("%s", linkedList->telNo); /// See above
if (i != listSize - 1) {
linkedList->next = (struct Entry*)malloc(sizeof(struct Entry)); /// See above
linkedList = linkedList->next; /// Again - changed here to use pointers!
}
else {
linkedList->next = NULL; /// See above
}
}
}
void print(struct Entry* linkedList) /// Pointer (see above)
{
int nr = 1;
printf("\nTelephone book is:\n");
while (current != NULL) { /// You need to check if "current" is not NULL before trying to access any of its members...
// while (current.name != NULL) {
printf("%d: %s has tel. No.\t%s\n", nr, current->name, current->telNo); /// Same "." to "->" changes as before
current = current->next; /// And again!
nr++;
}
}
There are other ways to 'fix' your code (and also improve it): for example, you never use the argument you pass to print, relying instead on the 'global' variable I mentioned earlier.