C fifo linked list char push - c

Question

I'm currently trying to understand fifo linked list and found example here Example , and I'm trying to input char instead of int
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
struct Node
{
char Data;
struct Node* next;
}*rear, *front;
void delQueue()
{
struct Node *temp, *var=rear;
if(var==rear)
{
rear = rear->next;
free(var);
}
else
printf("\nQueue Empty");
}
void push(char *value)
{
struct Node *temp;
temp=(struct Node *)malloc(sizeof(struct Node));
temp->Data=value;
if (front == NULL)
{
front=temp;
front->next=NULL;
rear=front;
}
else
{
front->next=temp;
front=temp;
front->next=NULL;
}
}
void display()
{
struct Node *var=rear;
if(var!=NULL)
{
printf("\nElements are as: ");
while(var!=NULL)
{
printf("\t%d",var->Data);
var=var->next;
}
printf("\n");
}
else
printf("\nQueue is Empty");
}
int main()
{
int i=0;
char ch;
front=NULL;
printf(" \n1. Push to Queue");
printf(" \n2. Pop from Queue");
printf(" \n3. Display Data of Queue");
printf(" \n4. Exit\n");
while(1)
{
printf(" \nChoose Option: ");
//scanf("%d",&i);
ch = getchar();
switch(ch)
{
case '+':
{
char value[20];
printf("\nEnter a valueber to push into Queue : ");
scanf("%s", value);
push(value);
printf("%s",value);
display();
break;
}
case '-':
{
delQueue();
display();
break;
}
case '*':
{
display();
break;
}
case '$':
{
exit(0);
}
default:
{
printf("\nwrong choice for operation");
}
}
}
}
I can't understand this warning on line 26: warning: assignment makes integer from pointer without a cast
I can input text, for example: "Hello world", but when I want to display it, it is showed as "-9".
I am really confused.

Answer 1

Data is defined as char, but your assigning a char * (char pointer) to it. That creates the warning, and surely won't work as you expect.

Answer 2

In your Node struct, the value is of type char, but you're assigning a char* to it instead. This is why you get the warning, and why printing things doesn't come out right.