[Solved] Why does the java accepts integer with a ‘+’ sign and how to not accept integer input with a ‘+’ sign

Question

You are reading an int using Scanner.nextInt(): as described in the documentation, this uses Integer.parseInt to read the number; and that method explicitly states that it accepts a leading + sign:

The characters in the string must all be digits of the specified radix (as determined by whether Character.digit(char, int) returns a nonnegative value), except that the first character may be an ASCII minus sign ‘-‘ (‘\u002D’) to indicate a negative value or an ASCII plus sign ‘+’ (‘\u002B’) to indicate a positive value

And once you’ve read that number, there’s no way to distinguish the fact that you entered 123 or +123, because there’s no difference in the value. So, you’ve lost the + even before you convert the int to a String.

To capture this, you need to read the String first, and convert that to an int:

String sAmount= scan.next();
nAmount = Integer.parseInt(sAmount);

This preserves the + sign in sAmount, because there is no reason to strip it away. Note that it will fail if sAmount can’t actually be parsed as an int.

Accepted Answer

You are reading an int using Scanner.nextInt(): as described in the documentation, this uses Integer.parseInt to read the number; and that method explicitly states that it accepts a leading + sign:

The characters in the string must all be digits of the specified radix (as determined by whether Character.digit(char, int) returns a nonnegative value), except that the first character may be an ASCII minus sign ‘-‘ (‘\u002D’) to indicate a negative value or an ASCII plus sign ‘+’ (‘\u002B’) to indicate a positive value

And once you’ve read that number, there’s no way to distinguish the fact that you entered 123 or +123, because there’s no difference in the value. So, you’ve lost the + even before you convert the int to a String.

To capture this, you need to read the String first, and convert that to an int:

String sAmount= scan.next();
nAmount = Integer.parseInt(sAmount);

This preserves the + sign in sAmount, because there is no reason to strip it away. Note that it will fail if sAmount can’t actually be parsed as an int.