Integer.parseInt(“9999999990”); [duplicate]

Answer 1

An int can only have a maximum value of 2147483647, so if you try to parse a number larger than that you'll get an exception (since it's not a valid int.)

Long.parseLong() will handle larger numbers, though bear in mind it still has an upper limit (2^63-1). If you really need a type without an upper limit, you'll have to use BigInteger (which has a constructor that takes a string.)

(You can find more information on all the primitive data types and their maximum values here.)

Answer 2

Integer.parseInt will throw an exception when what it's parsing can't be represented as an int. The first example is almost 10 billion, which is larger than the largest possible int, which is a little over 2 billion.

Integer.parseInt(String) delegates to Integer.parseInt(String, 10), the version that takes a radix, and those Javadocs state:

An exception of type NumberFormatException is thrown if any of the following situations occurs:

• The first argument is null or is a string of length zero.
• The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
• Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D') or plus sign '+' ('\u002B') provided that the string is longer than length 1.
• The value represented by the string is not a value of type int.

(emphasis mine)

If you need it parsed, you can use Long.parseLong, which will handle larger numbers.