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Integer.parseInt ("9999999990");

I am getting NumberFormatException for line 9999999990, but not when I use 1111111110 when I call Integer.parseInt on this Line. Let me know what is wrong.

String str="9999999990";
  int f = Integer.parseInt("2147483647");// No Exception here
        int x =Integer.parseInt(str);   // Exception is thrown here
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2 answers


Integer.parseInt

will throw an exception if its parse cannot be represented as int

. The first example is almost 10 billion, which is more than the maximum possible int

, which is just over 2 billion.

Integer.parseInt(String)

delegates Integer.parseInt(String, 10)

, the version that takes the root, and the state of the Javadocs:

A NumberFormatException is thrown when one of the following situations occurs:

  • The first argument is null or a zero-length string.
  • The radius is either less than Character.MIN_RADIX or greater than Character.MAX_RADIX.
  • Any character in the string is not a digit of the specified radius, except that the first character can 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 an int .


(emphasis mine)

If you need it, you can use Long.parseLong

one that will handle large numbers.

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An int

can only have a maximum value of 2147483647, so if you try to parse a number greater than that, you will get an exception (since it is not valid int

.)

Long.parseLong()

will handle large numbers, but remember that it still has an upper limit (2 ^ 63-1). If you really want a type with no upper limit, you will need to use BigInteger

(which has a constructor that takes a string).



(More information on all primitive data types and their maximum values ​​can be found ).

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