Why Is 09 "Too Large" of an Integer Number

Why is 09 too large of an integer number?

Numbers beginning with 0 are considered octal – and 9 is not an octal digit (but (conventionally) 0-7 are).

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Hexadecimal literals begin with 0x, e.g. 0xA.

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Up until Java 6, there was no literal notation for binary and
you'll had to use something like

int a = Integer.parseInt("1011011", 2);

where the second argument specifies the desired base.

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Java 7 now has binary literals.

In Java SE 7, the integral types (byte, short, int, and long) can also be expressed using the binary number system. To specify a binary literal, add the prefix 0b or 0B to the number.

Integer Number too large in java int for the value 08 , but when I pass just 8 or 11,12,13 it is just fine

It's because, in Java, numbers that start with a leading 0 (like your 08) are treated as octal (base 8). And there is no 8 in octal.

(By definition octal only uses the digits 0-7)

As an experiment, you can try 07 or 011 and see that they work, or try 08 or 09 and see that they don't work.

Integer number too large in Long type variable

08 and 00008 are parsed as octal numbers, in which 8 and 9 are invalid digits. Remove the leading zeroes.

Integer number too large (android / java)

In Java, numbers with a leading 0 are treated as octal numbers. Octal numbers are base 8 and use the digits 0 through 7, while the digits 8 and 9 are not valid.

To declare a decimal constant, use:

private static final int REQUEST_CODE = 909;

Java long number too large error?

All literal numbers in java are by default ints, which has range -2147483648 to 2147483647 inclusive.

Your literals are outside this range, so to make this compile you need to indicate they're long literals (ie suffix with L):

long min = -9223372036854775808L;
long max = 9223372036854775807L;

Note that java supports both uppercase L and lowercase l, but I recommend not using lowercase l because it looks like a 1:

long min = -9223372036854775808l; // confusing: looks like the last digit is a 1
long max = 9223372036854775807l; // confusing: looks like the last digit is a 1

Java Language Specification for the same

An integer literal is of type long if it is suffixed with an ASCII letter L or l (ell); otherwise it is of type int (§4.2.1).

Integer number too large

You need to use 4545454545l or 4545454545L to qualify it as long. Be default , 4545454545 is an int literal and 4545454545 is out of range of int.

It is recommended to use uppercase alphabet L to avoid confusion , as l and 1 looks pretty similar

You can do :

if(Long.valueOf(4545454545l).equals(Long.parseLong(morse)) ){
     System.out.println("2");
}

OR

if(Long.parseLong(morse) == 4545454545l){
   System.out.println("2");
}

As per JLS 3.10.1:

An integer literal is of type long if it is suffixed with an ASCII letter L or l (ell); otherwise it is of type int (§4.2.1).

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