This question already has answers here:
Why are integer literals with leading zeroes interpreted strangely?
(8 answers)
Closed 7 years ago.
I have a code like this, but I don't know why result variable have false value after execution the code
int x = 234;
boolean result = (x<0250);
and also why the following code doesn't work properly?
System.out.println(0250);
it prints 168 !! why?!
Integer literals starting with 0 are octal (base 8) not decimal (base 10).
Your options are
hexadecimal = 0x0C;
decimal = 12;
octal = 014;
binary = 0b1100;
A number that looks like 0x followed by an integer (using the digits 0-9 and A-F) is a hexadecimal (base 16) number.
A number that looks like 0 followed by an integer (using the digits 0-7) is an octal (base 8) number.
Related
This question already has answers here:
Convert hex string to int
(9 answers)
Closed 2 years ago.
I need to convert hexadecimal value which is OX12 to decimal in JAVA,
as I knew hexadecimal is based 16 (1..9, a..f), but in this case, I don't know how to convert with that value.
Can someone help me?
Thanks so much
If the number is a String, you can use the parseInt method of the Integer class with first argument "12" (the number in hex) and second argument 16, the radix, or base, of the number
int number = Integer.parseInt("12", 16);
System.out.println(number);
// output is 18
If the number is not a String, you can simply do the following:
int number = 0x12;
System.out.println(number);
// output is 18
I have the code to write a program that people enter a decimal value and i have to convert it to Hex, Octal, Binary, and Character values. Everything but Character value will print out in java eclipse. Simple program but confused.
Try something like this:
int i=65;
System.out.println((char)i);
Convert the integer to string, and take it digit by digit.
Then add 48 or (30 Hex) to each digit to get its ASCII code.
e.g. int(5) will be (5 + 48) = 53
e.g. int(16) will be (6 + 48) = 54, and (1 + 48) = 49
I hope this solves it.
This question already has answers here:
Change sign using bitwise operators
(3 answers)
Closed 7 years ago.
I need to negate a number using shift operator.
Example:
If number = 5 then negation of 5 should be -5
If number =-5 then negation of -5 should be 5.
Not with the shiftoperator but there is other bitwise operator using them you can do this
int i = 10;
i = (~i)+1;
System.out.println(i);
i = (~i)+1;
System.out.println(i);
i = (~i)+1;
System.out.println(i);
i = (~i)+1;
System.out.println(i);
result
-10
10
-10
10
BTW it is example of Two's complement and used for binary signed number representations
This question already has answers here:
why is not (123 == 0123) in java?
(3 answers)
Closed 9 years ago.
I am confused why this is and I cannot seem to find an answer why. This is from the assignment:
x=1, y=2, z=3;
z=(int)(x/y*3.0+z*012);
System.out.printf("%d %d %d", x, y, z);
Answer is :
1 2 30; << from eclipse
How I arrived here:
(1/2) = 0 * 3.0 = 0 + (z*012)= 30. I wanted to say 36 but I guess it is 30 according to the IDE.
012 is octal number not decimal which decimal value is 10.
z=(int)(x/y*3.0+z*012);
is equals -
z=(int)(1/2*3.0+3*10);
For reference
Numeric starts with 0 is octal number.
Numeric starts with 0x is hexadecimal number.
Numeric starts with 0b or OB is binary number.(Since Java edition 7 - Binary Literals)
In Java and several other languages, an integer literal beginning with 0 is interpreted as an octal (base 8) quantity. Here 012 is an octal number which has a decimal value f 10
So your multiplication will come like
z = (int) (1/2 * 3.0 + 3 * 10);
From JLS
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.
012 is an octal, because it starts with 0:
012 = (0 * 8^2) + (1 * 8^1) + (2) = 10
Therefore:
012 * 3 = 10 * 3 = 30
Notes:
Remember that an octal is a number in base 8 (decimal is base 10), so it can't have digits larger or equal to 8.
Similarly, hexadecimal numbers starts with 0x, for example: 0x12 = 1*16 + 2 = 18
See the JLS:
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.
So,
012 = 0 * 82 + 1 * 81 + 2 * 80 = 10
In Java 7, you can use underscores in numeric literals which might help you interrupting the value.
I am trying to write following code.but it gives me error kindly help me.
int six=06;
int seven=07;
int abc=018;
int nine=011;
System.out.println("Octal 011 ="+nine);
System.out.println("octal O18 =" + abc);
why i cant give 018 and 019 to variable.i can give value 020 and 021 to variable.
Why this happen? what's the reason behind this Kindly tell me.
I got Following error
integer number too large: 018
int eight=018;
Octal is base-8 number system, so it means digit can be from 0 to 7, you can't use digit 8 (and 9 too) in octal number system.
// Decimal declaration and possible chars are [0-9]
int decimal = 495;
// HexaDecimal declaration starts with 0X or 0x and possible chars are [0-9A-Fa-f]
int hexa = 0X1EF;
// Octal declaration starts with 0 and possible chars are [0-7]
int octal = 0757;
// Binary representation starts with 0B or 0b and possible chars are [0-1]
int binary = 0b111101111;
If the number is string format then you can convert it into int using the below
String text = "0b111101111";
int value = text.toLowerCase().startsWith("0b") ? Integer.parseInt(text.substring(2), 2)
: Integer.decode(text);
why i cant give 018 and 019 to variable.
Because an integer literal prefixed with 0 is treated as octal, and '8' and '9' aren't valid octal digits.
From section 3.10.1 of the JLS:
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.
Trying to use '8' in an octal number is like trying to use 'G' in hex... it's simple not part of the set of symbols used in that base.
Octal numbers (base 8) can only use the following figures: 01234567. The same way that decimal numbers (base 10) can only use 0123456789.
So in octal representation, 17 + 1 is 20.
The prefix 0 indicates octal(8 base)(digits 0-7).
public class MainClass{
public static void main(String[] argv){
int intValue = 034; // 28 in decimal
int six = 06; // Equal to decimal 6
int seven = 07; // Equal to decimal 7
int eight = 010; // Equal to decimal 8
int nine = 011; // Equal to decimal 9
System.out.println("Octal 010 = " + eight);
}
}
why i cant give 018 and 019 to variable.i can give value 020 and 021 to variable.
The leading zero signifies an octal literal. However, 8 and 9 are not valid octal digits. This makes 018 and 019 invalid.
When an integer literal starts with 0 in Java, it's assumed to be in octal notation. The digits 8 and 9 are illegal in octal—the digits can range only between 0 and 7.
Because it's octal, an octal number has 8 digits which spans from 0 to 7 inclusive. For the same reason 12 would be an invalid binary number.
You need at least base 9 to have 18 and a normal decimal base for 19.
For your query.....you assign an invalid value to the variable....your assigned value is started with 0(zero) ..which means that you are assigning an octal value to the variable and when you assign a value higher then 7 such as 018 in your case...the value excedes the range of octal variables and hence show an error...so try entering simply 18 so it would take it as an integer rather than an octal variable data type...