Thread: Question regarding stacks

I need a little explanation on one of the lines in this program. What does (what-'0') do? Thanks in advance.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define PUSH 1
#define POP 2
struct LLNode{
	char key;
	struct LLNode *next;
};
struct LLNode *initLLNode(char val);
char empty(struct LLNode *head, struct LLNode *tail);
void push(char val, struct LLNode *head);
char pop(struct LLNode *curr);
int main(void){
	char value, what;
	struct LLNode *head;
	struct LLNode *tail;	
	head=initLLNode(NULL);
	tail=initLLNode(NULL);

 	head->next=tail;
	printf("options:\n\t[%d]push\n\t[%d]pop\n\t[0]end\n",
		 PUSH,POP);
	do{
		printf("\n\t\t\t\t[%d][%d][0]>>",PUSH,POP);
		while(isspace(what=getchar()));
		switch(what-'0'){
			case PUSH :
				printf("\t\t\t\tPush what? ");
				while (isspace(value=getchar()));
				push(value,head);
				printf("'%c' pushed\n",value);
				break;

			case POP :
				if(empty(head,tail))printf("Stack Empty\n");
				else{
					value = pop(head);
					printf("'%c' popped\n", value);
				}
				break;
			default:	break;
		}
	}while(what-'0');
	printf("program finished\n");
	return(0);
}
struct LLNode *initLLNode(char val){
	struct LLNode *temp;
	temp=(struct LLNode *)malloc(sizeof(struct LLNode));
	temp->key=val;
	temp->next=NULL;
	return(temp);
}
char empty(struct LLNode *head, struct LLNode *tail){
/* returns 1 if empty, 0 otherwise */
	if(head->next==tail)return(1);
	else return(0);
}

void push(char val, struct LLNode *head){
	struct LLNode *temp;
	temp=initLLNode(val);
	temp->next=head->next;
	head->next=temp;
	return;
}



char pop(struct LLNode *curr){
	struct LLNode *temp;
	char val;
	val=curr->next->key;
	temp=curr->next;
	curr->next=curr->next->next;
	free(temp);
	return(val);
}