Number format exception in parse int

Question

public static void main(String[] args)throws IOException {
  String s ="12312a";
  int x = Integer.parseInt(s);
  System.out.println (x+2);
}

and all what I've got is :

Exception in thread "main" java.lang.NumberFormatException: For input string: "12312a"

any hints ?

"12312a" its not a valid number because a is there. "12312" is integer that can parse but a is Coming Under String so you cant . — Anand Dwivedi, Jan 06 '16 at 10:33

score 5 · Accepted Answer · answered Jan 06 '16 at 10:36

5

Perhaps you meant

int x = Integer.parseInt("12312a", 16);

answered Jan 06 '16 at 10:36

Reimeus

ThisaruG · Answer 2 · 2016-01-06T11:21:51.713

2

You can't parse a String to an int if the String isn't a number.

eg:

This will compile

String num = "3245";
int x = Integer.parseInt(num);

This will not:

String s ="12312a";
int x = Integer.parseInt(s);

Remove the a from your String.

If you want to parse it to the hexadecimal value, use

int x = int x = Integer.parseInt(s, 16);

This will parse it to base 16 number.

edited Jan 06 '16 at 11:21

answered Jan 06 '16 at 10:32

ThisaruG

score 2 · Answer 3 · answered Jan 06 '16 at 10:37

If you try to parse a String which is not a number you get a java.lang.NumberFormatException.

Maybe you want to parse a hexadecimal value, then you could use:

public static void main(String[] args) throws IOException {
 String s ="12312a";
 int x = Integer.parseInt(s,16);
 System.out.println (x+2);
}

I hope it helps. Have a nice day.

3 Answers3