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Accepted
D. Albarracín
Asked: 2020-03-12 10:15:00 +0800 CST 2020-03-12 10:15:00 +0800 CST

'The literal of type int is out of range' with small integers (2 digits)

  • 772

I create a variable of type int and I give it a value of, for example, "06" or some other value with a "0" in front, and this should show an error because if it is an integer it cannot have a zero to the left, but everything works fine. On the other hand, if we give it the value: "08" or "09", you get an error indicating that it is out of range.

The error should mark it with the entire sequence from "01" to "09", why only with "08" and "09"?

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public static void main( String[] args ) {
  int test = 08;
}

The literal 08 of type int is out of range

java

2 Answers

  1. Best Answer

    08, contrary to what you think, is a literal number in octal format . Java interprets as numbers in that format all those that start with a 0.

    In this format, the individual digits go from 0to7 .

    Hence the error: 08 is not a valid octal number , the same as 09. If they are correct, for example, 05' o03`.

    • 9

  2. Googling this: literal int java leading zeros (since the question sounded interesting to me), I found this answer in English .

    There it clarifies that if you put a leading 0, the value is taken as an octal. And octals are numbers from 0 to 7.

    Therefore, your 6 is an octal 6, not a 6 in base ten.

    Jon , in a later reply , provides us with a link to the java documentation .

    There he clarifies the answer a little more, saying that a single 0, or the numbers from 1 to 9 are taken as decimals, and any number that begins with 0 is taken as octal.

    As a curiosity, it clarifies that a 00 is a 0 in octal, as if it were really very different from a 0 in decimal.

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