Why does x = 12345678901234567890 return "x is less than y" when y = 1?

#include <stdio.h>

int main(void){

int x;
int y;
printf("What's x? ");
scanf("%d",&x);

printf("What's y? ");
scanf("%d",&y);

if (x < y)
{
    printf("x is less than y\n");
}
else if (x > y)
{
    printf("x is greater than y\n");
}
else
{
    printf("x and y are equal");
}

return 0;
}

Input:
12345678901234567890
1

Output:
x is less than y

expecting program to return "x is greater than y" when x = 12345678901234567890 and y = 1

>Solution :

The %d conversion specifier requires the corresponding scanf argument to be a pointer to int. The decimal digit sequence 12345678901234567890 corresponds to a number that is very likely to be outside the range that your int (probably a 32-bit type) can represent. Supposing that it is indeed out of range, the behavior of your program is undefined when processing the input specified. That means anything can happen, as far as the language spec is concerned.

C23 7.23.6.2/10:

the result of the conversion is placed in the object pointed to by the
first argument following the format argument that has not already
received a conversion result. If this object does not have an
appropriate type, or if the result of the conversion cannot be
represented in the object, the behavior is undefined.

(Emphasis added.)