Object x = someApi(); if (x instanceof Integer) Note that if someApi() returns type Integer the only possibilities of something returned are: an Integer null In which case you can: if (x == null) { // not an Integer } else { // yes an Integer }
See the Integer class and the static parseInt() method: http://developer.android.com/reference/java/lang/Integer.html Integer.parseInt(et.getText().toString()); You will need to catch NumberFormatException though in case of problems whilst parsing, so: int myNum = 0; try { myNum = Integer.parseInt(et.getText().toString()); } catch(NumberFormatException nfe) { System.out.println(“Could not parse ” + nfe); }
The immediate problem is due to you using <= temp.length() instead of < temp.length(). However, you can achieve this a lot more simply. Even if you use the string approach, you can use: String temp = Integer.toString(guess); int[] newGuess = new int[temp.length()]; for (int i = 0; i < temp.length(); i++) { newGuess[i] = temp.charAt(i) – ‘0’; } … Read more
If input value can be in numeric form other than integer , check by if (x == (int)x) { // Number is integer } If string value is being passed , use Integer.parseInt(string_var). Please ensure error handling using try catch in case conversion fails.
public class Three { public static void main(String[] args) { Three obj = new Three(); obj.function(600851475143); } private Long function(long i) { Stack<Long> stack = new Stack<Long>(); for (long j = 2; j <= i; j++) { if (i % j == 0) { stack.push(j); } } return stack.pop(); } } When the code above … Read more
