Why does this code throw NumberFormatException?
int a = Integer.parseInt("1111111111111111111111111111111111111111");
How to get the value of int for that String?
3 Answers
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.
Comments
This is because the number string is pretty large for an int . Probably this requires a BigInteger .
Comments
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.
String(viewed from the perspective as anint) exceedsInteger.MAX_VALUE(which is 2^31).Stringis a not a validint.inthas a maximum value of 2,147,483,647. Are you trying to parse a binary representation instead? If you are, you've got 40 bits there, so it still won't fit (int only stores 32).