Why does this code throw NumberFormatException?
int a = Integer.parseInt("1111111111111111111111111111111111111111");
How to get the value of int for that String?
Why does this code throw NumberFormatException?
int a = Integer.parseInt("1111111111111111111111111111111111111111");
How to get the value of int for that String?
 
    
     
    
    The value that you're attempting to parse is much bigger than the biggest allowable int value (Integer.MAX_VALUE, or 2147483647), so a NumberFormatException is thrown.  It is bigger than the biggest allowable long also (Long.MAX_VALUE, or 9223372036854775807L), so you'll need a BigInteger to store that value.
BigInteger veryBig = new BigInteger("1111111111111111111111111111111111111111");
From BigInteger Javadocs:
Immutable arbitrary-precision integers.
 
    
    This is because the number string is pretty large for an int . Probably this requires a BigInteger .
 
    
    There is no integer value for that string. That's why it's throwing an exception. The maximum value for an integer is 2147483647, and your value clearly exceeds that.
