C programming why is junk printing before the value? - c

Question

I am making a linked list of objects. For some reason when i go to print the value of a value within the object (a char array) i'm getting junk printing before the value. Why is this happening and how do i get rid of it? Here is a part of my code:
int num = 0;
int input = 1;
int retval = 0;
struct PDB2 *llist;
char avalue[100] = "";
llist = (struct PDB2 *)malloc(sizeof(struct PDB2));
llist->data1[100] = NULL;
llist->next = NULL;
while(input != 0) {
printf("\n-- Menu Selection --\n");
printf("0) Quit\n");
printf("1) Insert\n");
printf("2) Delete\n");
printf("3) Search\n");
printf("4) Display\n");
scanf("%d", &input);
switch(input) {
case 0:
default:
printf("Goodbye ...\n");
input = 0;
break;
case 1:
printf("Your choice: `Insertion'\n");
printf("Enter the value which should be inserted: ");
scanf("%s", &avalue);
append_node(llist, avalue);
break;
case 2:
printf("Your choice: `Deletion'\n");
printf("Enter the value which should be deleted: ");
scanf("%s", &avalue);
delete_node(llist, avalue);
break;
case 3:
printf("Your choice: `Search'\n");
printf("Enter the value you want to find: ");
scanf("%s", &avalue);
if((retval = search_value(llist, avalue)) == -1)
printf("Value `%s' not found\n", avalue);
else
printf("Value `%s' located at position `%d'\n", avalue, retval);
break;
case 4:
printf("You choice: `Display'\n");
display_list(llist);
break;
} /* switch */
} /* while */
free(llist);
return(0);
}
void append_node(struct PDB2 *llist, char message[])
{
int x = 0;
while(llist->next != NULL)
{
llist = llist->next;
}
llist->next = (struct PDB2 *)malloc(sizeof(struct PDB2));
for(x = 0; x < 100; x++)
{
llist->next->data1[x] = message[x];
}
llist->next->next = NULL;
}
void display_list(struct PDB2 *llist)
{
while(llist->data1 == NULL)
{
llist = llist->next;
}
while(llist->next != NULL)
{
printf("%s ", llist->data1);
llist = llist->next;
}
printf("%s", llist->data1);
}

Answer 1

The char array in your initial node is uninitialized. Change
llist->data1[100] = NULL;
to
llist->data1[0] = '\0';
to stop it from being printed.

Answer 2

These two lines, together, are wrong:
char avalue[100] = "";
[...]
scanf("%s", &avalue);
The proper way to scanf a string is:
scanf("%s", avalue); // No Address-Of operator.

Answer 3

The important part of your code (based on your description of the problem) can be boiled down to this:
struct PDB2 {
char data[100];
...other fields...
};
struct PDB2 * llist = (struct PDB2 *)(malloc(sizeof(struct PDB2)));
llist->data[100] = NULL; /* This is wrong */
printf("%s", llist->data); /* This prints rubbish */
The marked line is wrong because:
NULL is meant to be used as a pointer value, not as a character. You probably mean to use '\0'.
The point seems to be to clear the string, so the correct method is
.
llist->data[0] = '\0';
Also, the call to scanf seems that it should be
char avalue[100];
scanf("%s", avalue);
since avalue is already of type char *. With the extra & you may be writing to an invalid location. However,
int input;
scanf("%d", &input);
is correct for integer input.