Basic Linked List operations in C - c

Question

I am creating a program to perform basic linked list operations. Right now i have wrote the code only for inserting the node at the front. I ran my program to see its working or not but the program is terminating after accepting the input for the node, then it prints the message after switch. It doesn't even pause for accepting my input for continuing the operations (just before end of main())
here is the code :
#include <stdio.h>
#include <stdlib.h>
struct linkedlist
{
int num;
struct linkedlist *next;
};
struct linkedlist *head = NULL;
void display();
void insertBeginning()
{
struct linkedlist *obj;
int no;
obj = (struct linkedlist *)malloc(sizeof(struct linkedlist));
if(obj == NULL)
{
printf("\n Overflow ");
}
else
{
printf("\n Enter the number = ");
scanf("%d", &no);
obj->num = no;
if(head == NULL)
{
head = obj;
obj->next = NULL;
}
else
{
obj->next = head;
head = obj;
}
}
}
void display ()
{
struct linkedlist *head2 = head;
while(head2 != NULL)
{
printf("%d ->",head2->num);
head2=head->next;
}
printf("NULL \n");
}
int main()
{
int choice;
char wish;
printf("\n 1. Insert at beginning");
printf("\n 2. Insert at end");
printf("\n 3. Insert in between");
printf("\n 4. Delete from front");
printf("\n 5. Delete from end");
printf("\n 6. Delete from in between");
printf("\n 7. Reverse");
printf("\n 8. Sort ascending");
printf("\n 9. Sort descending");
printf("\n 10.Swap alternate elements");
printf("\n 11.Display\n\n");
do
{
printf("\n Enter the option = ");
scanf("%d", &choice);
switch(choice)
{
case 1:
insertBeginning();
break;
case 2:
// insertEnd();
break;
case 3:
// insertInbetween();
break;
case 4:
// deleteFront();
break;
case 5:
// deleteEnd();
break;
case 6:
// deleteInbetween();
break;
case 7:
// Reverse();
break;
case 8:
// sortAsc();
break;
case 9:
// sortDesc();
break;
case 10:
// swap();
break;
case 11:
display();
break;
default:
printf("\n Wrong choice ");
}
printf("\n Do you wish to continue (y/n) = ");
scanf ("%c",&wish);
}while(wish == 'y' || wish =='Y');
return 0;
}

Answer 1

In your case, you have to change
scanf ("%c",&wish);
to
scanf (" %c",&wish);
because, if you don't include the leading white-space before the format specifier, it will consider the remaining \n (newline) which got generated and stored into the input buffer by pressing ENTER key after the first input. So, the second scanf() won't wait for the user input.

Answer 2

when calling scanf()
1) with a '%d' format specifier, the trailing newline, from where the user entered the number, will not be consume.
2) with a '%c' format specifier, leading white space, like a newline, will cause the scanf() to fail, leaving the parameter (wish) unchanged.
3) in the posted code, when the 'wish' does not contain a valid 'Y' or 'y' then the program exits.
I agree with the other poster, that adding a choice '0' for exiting would be a much better way than the separate call to scanf()

Answer 3

there is a new line after giving the input 'choice' which scan by the variable 'wish'. So we need to remove that newline ('\n').
So if you want the user to continue just use a getchar() before take the input wish. Its easy and simple.
printf("\n Do you wish to continue (y/n) = ");
getchar();
scanf ("%c",&wish);