Develop Reference

Why would 0042 sound 34 to int? [duplicate]

This prints 83
System.out.println(0123)
However this prints 123
System.out.println(123)
Why does it work that way?

A leading zero denotes that the literal is expressed using octal (a base-8 number).
0123 can be converted by doing (1 * 8 * 8) + (2 * 8) + (3), which equals 83 in decimal.
For some reason, octal floats are not available.
Just don't use the leading zero if you don't intend the literal to be expressed in octal.
There is also a 0x prefix which denotes that the literal is expressed in hexadecimal (base 16).

Because integer literals starting with 0 are treated as octal numbers.
See section 3.10.1 of the JLS

Try this:
public static String leftPad(int n, int padding) {
return String.format("%0" + padding + "d", n);
}
leftPad(5, 3); // return "005"
leftPad(15, 5); // return "00015"
leftPad(224, 3); // return "224"
leftPad(0, 4); // return "0000"

first one printed as 83 because java takes 0123 as octal number and it prints decimal equivalent of that number.

The octal (leading 0) and hexadecimal (leading 0x) were inherited from C.
For comparison, try
System.out.println(0x123);

In Java integer literals with a leading zero are octal integers (base 8).
(1 * 8^2) + (2 * 8^1) + (3 * 8^0) = 83
So do not use any number leading with 0 if you don't want to treat it as an octal number.

0123 -> 83
1010L -> 1010
0101L -> 65
The numbers 1010L and 0101L are not in binary representation (just to avoid the confusion).
These numbers are in decimal representation.
Even as per the Regex patterns in Oracle docs,
\0n is the character with octal value 0n (0 <= n <= 7)
\xhh is the character with hexadecimal value 0xhh
Thus, your number 0101 be it in Integer or Long format is treated as an Octal representation of a number.
123 => 1 * 8^2 + 2 * 8^1 + 1 * 8^0 = 83
0101 => 1 * 8^2 + 0 * 8^1 + 1 * 8^0 = 64 + 0 + 1 = 65

printf will do it: http://java.sun.com/developer/technicalArticles/Programming/sprintf/
public class X
{
public static void main(final String[] argv)
{
System.out.printf("%04d", 123);
System.out.println();
}
}
You could also make it "%0" + size + "%d" if you wanted to vary the length... though if the lengths were common I'd probably make constants like "%04d", "%012d", etc...

Long starting with zero has wrong value [duplicate]

This prints 83
System.out.println(0123)
However this prints 123
System.out.println(123)
Why does it work that way?

A leading zero denotes that the literal is expressed using octal (a base-8 number).
0123 can be converted by doing (1 * 8 * 8) + (2 * 8) + (3), which equals 83 in decimal.
For some reason, octal floats are not available.
Just don't use the leading zero if you don't intend the literal to be expressed in octal.
There is also a 0x prefix which denotes that the literal is expressed in hexadecimal (base 16).

Because integer literals starting with 0 are treated as octal numbers.
See section 3.10.1 of the JLS

Try this:
public static String leftPad(int n, int padding) {
return String.format("%0" + padding + "d", n);
}
leftPad(5, 3); // return "005"
leftPad(15, 5); // return "00015"
leftPad(224, 3); // return "224"
leftPad(0, 4); // return "0000"

first one printed as 83 because java takes 0123 as octal number and it prints decimal equivalent of that number.

The octal (leading 0) and hexadecimal (leading 0x) were inherited from C.
For comparison, try
System.out.println(0x123);

In Java integer literals with a leading zero are octal integers (base 8).
(1 * 8^2) + (2 * 8^1) + (3 * 8^0) = 83
So do not use any number leading with 0 if you don't want to treat it as an octal number.

0123 -> 83
1010L -> 1010
0101L -> 65
The numbers 1010L and 0101L are not in binary representation (just to avoid the confusion).
These numbers are in decimal representation.
Even as per the Regex patterns in Oracle docs,
\0n is the character with octal value 0n (0 <= n <= 7)
\xhh is the character with hexadecimal value 0xhh
Thus, your number 0101 be it in Integer or Long format is treated as an Octal representation of a number.
123 => 1 * 8^2 + 2 * 8^1 + 1 * 8^0 = 83
0101 => 1 * 8^2 + 0 * 8^1 + 1 * 8^0 = 64 + 0 + 1 = 65

printf will do it: http://java.sun.com/developer/technicalArticles/Programming/sprintf/
public class X
{
public static void main(final String[] argv)
{
System.out.printf("%04d", 123);
System.out.println();
}
}
You could also make it "%0" + size + "%d" if you wanted to vary the length... though if the lengths were common I'd probably make constants like "%04d", "%012d", etc...

Trying to print a casted array of ints into chars,but I get message "integer number too large" [duplicate]

This prints 83
System.out.println(0123)
However this prints 123
System.out.println(123)
Why does it work that way?

A leading zero denotes that the literal is expressed using octal (a base-8 number).
0123 can be converted by doing (1 * 8 * 8) + (2 * 8) + (3), which equals 83 in decimal.
For some reason, octal floats are not available.
Just don't use the leading zero if you don't intend the literal to be expressed in octal.
There is also a 0x prefix which denotes that the literal is expressed in hexadecimal (base 16).

Because integer literals starting with 0 are treated as octal numbers.
See section 3.10.1 of the JLS

Try this:
public static String leftPad(int n, int padding) {
return String.format("%0" + padding + "d", n);
}
leftPad(5, 3); // return "005"
leftPad(15, 5); // return "00015"
leftPad(224, 3); // return "224"
leftPad(0, 4); // return "0000"

first one printed as 83 because java takes 0123 as octal number and it prints decimal equivalent of that number.

The octal (leading 0) and hexadecimal (leading 0x) were inherited from C.
For comparison, try
System.out.println(0x123);

In Java integer literals with a leading zero are octal integers (base 8).
(1 * 8^2) + (2 * 8^1) + (3 * 8^0) = 83
So do not use any number leading with 0 if you don't want to treat it as an octal number.

0123 -> 83
1010L -> 1010
0101L -> 65
The numbers 1010L and 0101L are not in binary representation (just to avoid the confusion).
These numbers are in decimal representation.
Even as per the Regex patterns in Oracle docs,
\0n is the character with octal value 0n (0 <= n <= 7)
\xhh is the character with hexadecimal value 0xhh
Thus, your number 0101 be it in Integer or Long format is treated as an Octal representation of a number.
123 => 1 * 8^2 + 2 * 8^1 + 1 * 8^0 = 83
0101 => 1 * 8^2 + 0 * 8^1 + 1 * 8^0 = 64 + 0 + 1 = 65

printf will do it: http://java.sun.com/developer/technicalArticles/Programming/sprintf/
public class X
{
public static void main(final String[] argv)
{
System.out.printf("%04d", 123);
System.out.println();
}
}
You could also make it "%0" + size + "%d" if you wanted to vary the length... though if the lengths were common I'd probably make constants like "%04d", "%012d", etc...

Integer massive [duplicate]

This prints 83
System.out.println(0123)
However this prints 123
System.out.println(123)
Why does it work that way?

A leading zero denotes that the literal is expressed using octal (a base-8 number).
0123 can be converted by doing (1 * 8 * 8) + (2 * 8) + (3), which equals 83 in decimal.
For some reason, octal floats are not available.
Just don't use the leading zero if you don't intend the literal to be expressed in octal.
There is also a 0x prefix which denotes that the literal is expressed in hexadecimal (base 16).

Because integer literals starting with 0 are treated as octal numbers.
See section 3.10.1 of the JLS

Try this:
public static String leftPad(int n, int padding) {
return String.format("%0" + padding + "d", n);
}
leftPad(5, 3); // return "005"
leftPad(15, 5); // return "00015"
leftPad(224, 3); // return "224"
leftPad(0, 4); // return "0000"

first one printed as 83 because java takes 0123 as octal number and it prints decimal equivalent of that number.

The octal (leading 0) and hexadecimal (leading 0x) were inherited from C.
For comparison, try
System.out.println(0x123);

In Java integer literals with a leading zero are octal integers (base 8).
(1 * 8^2) + (2 * 8^1) + (3 * 8^0) = 83
So do not use any number leading with 0 if you don't want to treat it as an octal number.

0123 -> 83
1010L -> 1010
0101L -> 65
The numbers 1010L and 0101L are not in binary representation (just to avoid the confusion).
These numbers are in decimal representation.
Even as per the Regex patterns in Oracle docs,
\0n is the character with octal value 0n (0 <= n <= 7)
\xhh is the character with hexadecimal value 0xhh
Thus, your number 0101 be it in Integer or Long format is treated as an Octal representation of a number.
123 => 1 * 8^2 + 2 * 8^1 + 1 * 8^0 = 83
0101 => 1 * 8^2 + 0 * 8^1 + 1 * 8^0 = 64 + 0 + 1 = 65

printf will do it: http://java.sun.com/developer/technicalArticles/Programming/sprintf/
public class X
{
public static void main(final String[] argv)
{
System.out.printf("%04d", 123);
System.out.println();
}
}
You could also make it "%0" + size + "%d" if you wanted to vary the length... though if the lengths were common I'd probably make constants like "%04d", "%012d", etc...

How does a leading zero change a numeric literal in Java?

My friends and I have puzzled over this statement in Java after seeing it and the answer.
How does this work?
System.out.printf("%d", 077);
equals 63?

077 is an octal number which equals 7 x 81 + 7 x 80 which is 63 decimal. To display the octal value you could do
System.out.printf("%o", 077);

When you define an literal integer number with a 0 prefix, the compiler will treat it as an integer base 8. (Octal).
Check at this post http://rodrigosasaki.com/2013/06/10/number-literals-in-java/
So, 77 value in Octal base is actually 63 in Decimal base.

077 = 7 * 8^0 + 7 * 8^1 = 63;
0123 = 3 * 8^0 + 2 * 8^1 + 1 * 8^2 = 3 + 16 + 64 = 83;
First 0 means octal value.
0x77 - is hex val.