java - What's the difference when using numeric literal in termination expression of a for statement? -

why piece of code:

string value = joptionpane.showinputdialog("enter x"); //input = 100 int x = integer.parseint(value); double result = 1;  (int = 1; <= x; i++) //used variable "x" here {     result += (x * 1.0) / fact(i);     x *= x; }  public static int fact(int n) {     int fact = 1;     (int = 1; <= n; i++) {         fact *= i;     }     return fact; } 

work differently one?

string value = joptionpane.showinputdialog("enter x"); //input = 100 int x = integer.parseint(value);    double result = 1;  (int = 1; <= 100; i++) //and here used value "100" {     result += (x * 1.0) / fact(i);     x *= x; }  public static int fact(int n) {     int fact = 1;     (int = 1; <= n; i++) {         fact *= i;     }     return fact; } 

the change made using value 100 instead of using variable x in termination expression!

when run first code, get:

9.479341033333334e7 

however, second 1 get

nan 

why?

the difference between 2 snippets this:

for (int = 1; <= x; i++) { 

vs.

for (int = 1; <= 100; i++) { 

in first case, x gets larger every time! eventually, stop when x overflows , becomes 0, sooner in second case. explanation why results in 0 instead of other random number, see: why multiplication integer overflow result in zero?

in second case, when i = 34, fact(n) return 0, double division (0 * 1.0) /0 results in nan. double, when added nan, becomes nan, why second snippet results in nan. see: in java, nan mean?