java - Why does Scanner#nextInt inside for loop keep throwing an exception? -

java - Why does Scanner#nextInt inside for loop keep throwing an exception? -

i have been learning java , have little uncertainty code :

class apple { public static void main(string[] args) { int[] num = new int[3]; scanner input = new scanner(system.in); (int = 0; < num.length; i++) { seek { num[i] = input.nextint(); } grab (exception e) { system.out .println("invalid number..assigning default value 20"); num[i] = 20; } } (int = 0; < num.length; i++) { system.out.println(num[i]); } } }

i have written little programme handle exception, if user input not int throw exception , assign default value. if set scanner statement within loop, works fine, if take outside assign same value @ exception thrown i.e entering char rather int. if come in integers assign right values in array.

scanner input = new scanner(system.in);

i hope u guys have understood question.

scanner#nextint doesn't advance past input if fails parse integer, if maintain calling after failure, maintain trying parse same input again, throwing inputmismatchexception.

you can phone call scanner#next, ignoring string returns, in catch block skip invalid input:

try { num[i] = input.nextint(); } grab (exception e) { system.out .println("invalid number..assigning default value 20"); num[i] = 20; input.next(); }

java java.util.scanner