When I run this program, it outputs -43.
public class Main {
public static void main(String[] args) {
int a=053;
System.out.println(a);
}
}
Why is this? How did 053 turn into -43?
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5 Answers
The java tutorials http://download.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
int decVal = 26; // The number 26, in decimal
int octVal = 032; // The number 26, in octal <<== LOOK FAMILIAR?
int hexVal = 0x1a; // The number 26, in hexadecimal
int binVal = 0b11010; // The number 26, in binary
Yup... it's a gotcha!
Cheers. Keith.
Comments
It prints out 43, not -43. That is because if you write a number with a leading 0, it is an octal constant.
From here, http://download.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
int octVal = 032; // The number 26, in octal
1 Comment
corlettk
Nahhh... I'm just on the other side of the planet from the nearest SOF server... I have never supplied the first answer to any question, and I don't expect to... the question is (I suspect) a few minutes old before I even see it.
As http://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.1 says:
An octal numeral consists of an ASCII digit 0 followed by one or more of the ASCII digits 0 through 7 interspersed with underscores, and can represent a positive, zero, or negative integer.
And its format is:
OctalNumeral:
0 OctalDigits
0 Underscores OctalDigits
So, if you use
int octVal = 053;
or,
int octVal = 0_53;
both of them, you will get 43.
Comments
This from googling 'how to convert octal to decimal'
Octal to Decimal
- Start the decimal result at 0.
- Remove the most significant octal digit (leftmost) and add it to the result.
- If all octal digits have been removed, you're done. Stop.
- Otherwise, multiply the result by 8.
- Go to step 2.
So 0 + 5 = 5, 5 * 8 = 40, 40 + 3 = 43
