Why queue using two stacks not functioning? - c

Question

I'm trying to implement a queue using 2 stacks but my code isn't functioning.
Can you spot the error?
#include <stdio.h>
#include<stdlib.h>
#define MAX 5
typedef struct stack {
int top;
int arr[MAX];
} stack;
void enque(stack*s1, int ele) {
printf("entering");
push(&s1, ele);
printf("what a pain");
}
void push(stack*s, int ele) {
if (s->top == MAX - 1) {
printf("OVERFLOW");
} else {
s->arr[++s->top] = ele;
}
}
int deq(stack*s1, stack*s2) {
int x;
if (s1->top == -1 && s2->top == -1) {
printf("empty");
} else {
if (s2->top == -1) {
while (s1->top != -1) {
push(&s2, pop(&s1));
}
}
x = pop(&s2);
return x;
}
}
int pop(stack *s) {
if (s->top == -1) {
printf("UNDERFLOW");
} else {
return s->arr[s->top--];
}
}
void display(stack*s) {
printf("entered display");
int i;
for (i = 0; i <= s->top; i++) {
printf(" %d", s->arr[i]);
}
}
int main() {
int ch, ele, c;
stack s1, s2;
s1.top = -1, s2.top = -1;
do {
printf("1 - Enqueue2-deq3-display4-exit\n");
printf("Enter choice");
scanf("%d", &ch);
switch (ch) {
case 1:
printf("enter ele of ur choice");
scanf("%d", &ele);
enque(&s1, ele);
break;
case 2:
c = deq(&s1, &s2);
printf("%d", c);
break;
case 3:
display(&s1);
break;
case 4:
exit(0);
default:
printf("Wrong choice");
}
} while (ch != 5);
}

Answer 1

Can you spot the error?
The compiler can spot the error faster than Stack-overflow.
OP is not compiling with all warnings enabled or is using a weak compiler
Enable all warnings to save time.
int deq(stack*s1, stack*s2) and int pop(stack *s) both have the same problem.
A common waning in such cases is warning: control reaches end of non-void function [-Wreturn-type]
Make certain each function path returns a value.
int pop(stack *s) {
if (s->top == -1) {
printf("UNDERFLOW"); // Notice, no return
} else {
return s->arr[s->top--];
}
}
Also push(&s1, ele); used before declaration. This make for conflicting function signature. Declare or define push() before using. Likely undefined behavior (UB) and this makes the code unreliable.
I recommend to print an '\n' more often.
// printf("entering");
printf("entering\n");
// or
puts("entering"); // \n automatically added.

Answer 2

You are passing pointer address in deque and enque functions.
push(&s2, pop(&s1)); --> push(s2, pop(s1));
x = pop(&s2); --> x = pop(s2);
push(&s1, ele);-->push(s1, ele);
since s1 and s2 are received as pointers to deque and enque functions
Also consider changes suggested by #Mr.Chux by considering compiler warnings you could have resolved all the issues by yourself.