Unable to pop top element from the linked list - c

Question

I was trying to make a simple Linked List program, also when I'm trying to pop the first element from the list , it's not popping and it still remains the first element in the list, please help me resolve this error.
Here is the code:
#include <stdio.h>
#include <stdlib.h>
void create(stack *s){
if(s == NULL){
s = (stack*)malloc(sizeof(stack)*1);
(s->next)=NULL;
}
else{
stack *temp = (stack*)malloc(sizeof(stack)*1);
(temp->next)=s;
s=temp;
}
}
void push(stack *s, char x){
create(s);
(s->value)=x;
}
void isEmpty(stack *s){
if(s == NULL){
printf("List is Empty!\n");
}
else{
printf("List is not Empty!\n");
}
}
char pop(stack *s){
if(s == NULL){
isEmpty(s);
return -1;
}
char x=s->value;
s=(s->next);
return x;
}
int main(int argc , char* argv[]){
stack *s;
create(s);
char choice,data;
printf("Stack Created\n\n");
do{
printf("Choose Option: pUsh, pOp, pEek, iseMpty, getSize, eXit: \n");
scanf(" %c",&choice);
switch(choice){
case 'U':{
printf("Enter the element to be pushed: \n");
scanf(" %c",&data);
push(s, data);
break;
}
case 'O':{
data=pop(s);
if(data != NULL){
printf("Popped: %c\n", data);
}
break;
}
}
}while(1);
return 0;
}

Answer 1

The line s=s->next; has no effect because s is a local varaible. You need to return the new value of s or use pointers to modify the caller's version.

Answer 2

I changed the argument to pop() and push() from stack * to stack **, so that we can update the stack not a local variable. I removed create() as it is basically what happens during the push and I integrated it with that.
The rest is straightforward and I also added the free() call in pop(). Take a look:
#include <stdio.h>
#include <stdlib.h>
typedef struct stack{
int value;
struct stack *next;
}stack;
void push(stack **s, int x){
stack *temp = (stack*)malloc(sizeof(stack)*1);
temp->value = x;
temp->next = NULL;
if(*s == NULL){
*s = temp;
}else{
temp->next = *s;
*s=temp;
}
}
char pop(stack **s){
if(*s == NULL){
return -1;
}
char x=(*s)->value;
stack *tmp = *s;
*s=(*s)->next;
free(tmp);
return x;
}
int main(int argc , char* argv[]){
stack *s;
char choice,data;
printf("Stack Created\n\n");
do{
printf("Choose Option: pUsh, pOp, pEek, iseMpty, getSize, eXit: \n");
scanf(" %c",&choice);
switch(choice){
case 'U':{
printf("Enter the element to be pushed: \n");
scanf(" %c",&data);
push(&s, data);
break;
}
case 'O':{
data=pop(&s);
if(data != -1){
printf("Popped: %c\n", data);
} else {
printf("Stack is empty. nothing popped");
}
break;
}
}
}while(1);
return 0;
}

Answer 3

Here you have to use pointer to pointer if u want to pass pointer as an argument
char pop(stack **s)
{
**s=s->next;
}
this is a temporary solution u should also consider deleting the memory you allocated using the malloc function or else it will lead to memory leak
also pass the address of the pointer s while you call it in the pop function