How can I convert an int number from decimal to binary. For example:
int x=10; // radix 10
How can I make another integer has the binary representation of x, such as:
int y=1010; // radix 2
by using c only?
An integer is always stored in binary format internally -- saying that you want to convert int x = 10 base 10 to int y = 1010 base 2 doesn't make sense. Perhaps you want to convert it to a string representing the binary format of the integer, in which case you can use Integer.toBinaryString.
First thing you should understand is that a value is an abstract notion, that is not bounded to any representation. For example, if you have 20 apples, the number of apples will be the same regardless of the representation. So, dec("10") == bin("1010").
The value of an int reffers to this abstract notion of value, and it does not have any form until you with to print it. This means that the notion of base is important only for conversions from string to int and back.
String s = Integer.toBinaryString(10);
http://download.oracle.com/javase/1.5.0/docs/api/java/lang/Integer.html
Whether it's binary or decimal doesn't really have anything to do with the integer itself. Binary or decimal is a property of a physical representation of the integer, i.e. a String. Thus, the methods you should look at are Integer.toString() and Integer.valueOf() (the versions that take a radix parameter).
BTW, internally, all Java integers are binary, but literals in the source code are decimal (or octal).
Your question is a bit unclear but I'll do my best to try to make sense of it.
How can I make another integer has the binary representation of x such as: int y=1010 radix 2?
From this it looks like you wish to write a binary literal in your source code. Java doesn't support binary integer literals. It only supports decimal, hexadecimal and octal.
You can write your number as a string instead and use Integer.parseInt with the desired radix:
int y = Integer.parseInt("1010", 2);
But you should note that the final result is identical to writing int y = 10;. The integer 10 that was written as a decimal literal in the source code is identical in every way to one which was parsed from the binary string "1010". There is no difference in their internal representation if they are both stored as int.
If you want to convert an existing integer to its binary representation as a string then you can use Integer.toBinaryString as others have already pointed out.
Both integers will have the same interior representation, you can however display as binary via Integer.toBinaryString(i)
Use Integer.toBinaryString()
String y = Integer.toBinaryString(10);
Converting an integer to another base (string representation):
int num = 15;
String fifteen = Integer.toString(num, 2);
// fifteen = "1111"
Converting the string back into an integer
String fifteen = "1111";
int num = Integer.valueOf(fifteen, 2);
// num = 15
This covers the general case for any base. There's no way to explicitly assign an integer as binary (only decimal, octal, and hexadecimal)
int x = 255; // decimal
int y = 0377; // octal (leading zero)
int z = 0xFF; // hex (prepend 0x)
Related
I have this line of Java code that will throw NumberFormatException if the number represented as a String is above 2,147,483,647.
Because:
The int data type is a 32-bit signed two's complement integer. It has
a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647
Code Throwing the NumberFormatException:
String largeNumberAsAString = "9999999999";
Integer.toHexString(Integer.parseInt(largeNumberAsAString)); // NumberFormatException
How can I get the same functionality of theInteger.toHexString() with a String parameter and not an int parameter due to NumberFormatException?
Use BigInteger to avoid numeric limits of primitive int and long:
BigInteger x = new BigInteger("9999999999999999999999"); // Default radix is 10
String x16 = x.toString(16); // Radix 16 indicates hex
System.out.println(x16);
The class conveniently exposes a constructor that takes a String, which gets interpreted as a decimal representation of a number.
Demo.
If your input value can be arbitrarily large, then #dasblinkenlight's answer involving BigInteger is your best bet.
However, if your value is less than 263, then you can just use Long instead of Integer:
String dec = "9999999999";
String hex = Long.toHexString(Long.parseLong(dec));
System.out.println(hex); // 2540be3ff
Live demo.
Use Integer.parseUnsignedInt
When the number is above 2^31 but below 2^32, thus in the negative int range,
you can do:
int n = Integer.parseUnsignedInt("CAFEBABE", 16);
(I used hexadecimal here, as it is easier to see that above we are just in that range.)
However 9_999_999_999 is above the unsigned int range too.
Try this way:
String largeNumberAsAString = "9999999999";
System.out.println(Integer.toHexString(BigDecimal.valueOf(Double.valueOf(largeNumberAsAString)).intValue()));
PS: I tried searching many existing questions here on Stackoverflow, however it
only added chaos to my query!
10101
11100
11010
00101
Consider this as a sample Input, which I need to read as BinaryString one by one! Then I need to represent them as an Integer.
Obviously int x = 20 initializes x as a decimal Integer,
and I read from other questions that int y = 0b101 initializes y as a binary Integer.
Now, The question is: If I have a binaryString, how do I cast it into an int like with a preceding 0b . My objectives following this is to perform bit level OR and XOR operations.
ie from the above input, I need to do 10101 ^ 11100
Let me know if this is the right way to approach a problem like this, Thanks!
If I have understood your question correctly, you want to know how to represent Binary Strings as Integer.
Well, you can perform this task for conversion of Binary String to Integer:
String input = "0b1001";
String temp = input.substring(2);
int foo = Integer.parseInt(temp, 2);
Alternately, to switch back :
String temp = Integer.toBinaryString(foo);
from the above input, I need to do 10101 ^ 11100.
You can achieve the same using proper decimal representation of integer. If you want to re-invent the wheel, then
first convert the decimal representation of the given number to Binary String(using step 2);
then convert to integer value using step 1;
repeat steps 1 and 2 for the second number; and
finally, perform the XOR operation over them.
But, I don't see how it'll be performing/calculating differently. It'd still be stored as the same integer. It is just that you will get extra satisfaction(on your part) that the number was read as an integer and then converted to Binary representation of that number.
Try Integer.parseInt(String s, int radix). It will throw an exception if the binary string is invalid, so you might want to validate the input.
String binary = "10001";
int asInt = Integer.parseInt(binary, 2); // 17
int supports ^ (bitwise XOR) operator. All you have to do is convert your binary strings to int variables and perform the desired operations.
I have hexadecimal String eg. "0x103E" , I want to convert it into integer.
Means String no = "0x103E";
to int hexNo = 0x103E;
I tried Integer.parseInt("0x103E",16); but it gives number format exception.
How do I achieve this ?
You just need to leave out the "0x" part (since it's not actually part of the number).
You also need to make sure that the number you are parsing actually fits into an integer which is 4 bytes long in Java, so it needs to be shorter than 8 digits as a hex number.
No need to remove the "0x" prefix; just use Integer.decode instead of Integer.parseInt:
int x = Integer.decode("0x103E");
System.out.printf("%X%n", x);
I was playing a bit with numbers, and something interesting came upon me, which I don't quite understand.
public static void main(String[] args) {
int hexNumber = 0x7A;//decimal: 122 binary:0111 1010
int decNumber = 122;
int binNumber = 1111010;
System.out.println(hexNumber);//122
System.out.println(Integer.toString(hexNumber, 16)); //7a
System.out.println(Integer.toHexString(hexNumber)); //7a
System.out.println(Integer.toString(hexNumber, 2)); // 1111010
System.out.println(Integer.toBinaryString(hexNumber)); //1111010
System.out.println(hexNumber==binNumber);//false
System.out.println(hexNumber==decNumber);//true
System.out.println(decNumber==binNumber);//false
}
Why do I get "false" at #1 and #3? Doesn't change even if binNumber = 01111010;
Well, you can't directly store binary values in Java without any prefix.
binNumber isn't stored as the binary number 1111010; instead, it's stored as the decimal number 1111010.
This you have to store as int binNumber = Integer.parseInt("1111010", 2); or better yet int binNumber = 0b1111010;.
For octal:
int octalNo = 0177; //'0' is prefix
or
int octalNo = Integer.parseInt("0177", 8); //leading '0's are ignored
For hexadecimal:
int hexNo = 0x177; //'0x' is prefix
or
int hexNo = Integer.parseInt("0177", 16); //leading '0's are ignored
For more info, have a look at this.
You aren't creating the binary number as a binary one. You are creating it as a decimal one (base 10) that happens to only contain 0s and 1s.
To store 0111 1010 in Java 7 use the new binary literal (you can even use underscores for easier reading)
int binNumber = 0b0111_1010;
Because that's not the proper way to specify the binary number (and there is no such way in Java, apart from something like Integer.toBinaryString(122) which would give you a proper binary representation (returned as a String)).
Your number was interpreted as a "normal" decimal integer (if entered without leading 0) or as an integer in octal system (if entered with leading 0).
In Java versions before 7, you need to use this
int binNumber = Integer.parseInt("1111010", 2);
In 7 and up, you can use
int binNumber = 0b1111010;
With that change, your code works here (I get three true resuls).
The answer is clear. You have assigned to integers some diferent values.hexNumber is initialized to decimal value 122, even if the representation you use for that is hexadecimal. decNumber is initialized to decimal value 122, so when you compare hexNumber and decNumber you will get true because it's really the same value. Finally binNumber is initialized to the decimal value 1111010, so if you compare it with one of the other numbers you will get false.
In the following code intArray[i] stores RGB values of pixels in hex format(eg:0xffff0000) .... The method hsvToRgb() gives bak an integer value of RGB (eg:15777252) but i need back the rgb value in the original hex format after changes.
The second line gives me that but its a string ....What do i do to store this string value back into the array? ... please help me.
int disco = hsvToRgb(hsv);
hexColor = String.format("0x%06X", (0xffffff & disco));
intArray[i] = Integer.valueOf(String.valueOf(disco), 16);
There's no such thing as a "hex format" integer versus a "decimal format" integer. The bit/byte representation of the value is the same. For example, the decimal value 15,777,252 is the hex value 0xF0BDE4. (You can use Google to convert: search "15777252 in hex").
You can use the disco value directly. If you want to print it out in a hex representation, use Integer.toHexString().
Regarding the format. Think of it like this ... The computer represents the value as a series of bits. By way of example, let's pick a random number and represent it using 8 bits: 01110101. Using a bit string to represent bigger numbers would get very long very quickly, so hexadecimal is often used. The hex equivalent is: 65. By convention, we usually precede the value by 0x when it's in hex. That gives us 0x65. Non-programmers tend to deal more naturally in base 10 however (rather than base 16). The same number in base 10 is 101.
You can see this with some code:
final int value = 0x65; // we can declare it in hex
final int sameValue = 101; // or in decimal
System.out.println(value); // output in base 10; prints "101"
System.out.println(Integer.toHexString(value)); // output in base 16; prints "65"
System.out.println(Integer.toBinaryString(value)); // output in base 2; prints "1100101"
System.out.println(""+(value == sameValue)); // prints "true"