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Why doesn't Java support unsigned ints?
(17 answers)
Closed 6 years ago.
Byte value is between 0 and 255, very simple and straight forward. Java didn't think so however and decided that values to be between -128 and 128 and ruined my life and many others.
I just want to know what's the big idea? why everytime i need to get the unsigned byte value do I have to do this:
int byteValue = (int) javaByte & 0xFF;
Java supports only signed integers. byte is integer number as well so Java is just consistent here.
This is not case for C for example where char type doesn't have defined whether it's signed or unsigned. You need to explicitly tell. This is in my opinion much worse than saying that it's by default signed.
Anyway, the original intention in C for char was a string character where the unsigned made sense. In Java there is different type for string character - char.
Related
This question already has answers here:
Understanding Java data types
(2 answers)
How to represent 11111111 as a byte in java
(2 answers)
JAVA: why binary literal for byte with negative sign is being considered as integer type?
(5 answers)
Closed 1 year ago.
byte b = 0b1111_1111;
byte c = 0b111_1111;
byte d = 0b0111_1111;
The first line causes a compilation error: incompatible types. I also tried 0xFF and decimal, the same. Thanks.
And I think byte means cannot be larger than what 8 bits hold. 0b1111_1111 is 8 bits.
All basic integer types in Java are signed. To be exact, byte, short, int and long are signed big endian two complement integers. That means that they run from - 2n-1 to 2n-1 - 1 where n is the number of bits - 8 for a byte (Byte.SIZE in Java), and therefore the range is from -128 to 127.
The number you are using is always an integer; number literals always are. However, the integer is actually the value 0x000000FF in hexadecimals. That cannot be stored as it is higher than 127. You can however do an explicit conversion; your value does consist of 8 bits after all.
byte b = (byte) 0b1111_1111
works because it simply disregards the 24 bits from the left hand side of the value. Now if you print it you will get -1 as value though as 0b1111_1111 represents 1 in 8 bit two complement form.
If you need it to represent the integer value 255 then you need to perform the following little trick:
int i = b & 0xFF;
which will actually do the following:
First it will convert b to the value 0xFFFFFFFF as integer using sign extension as all calculations default to integer, and it still thinks that b represents -1 after all.
Then it performs a bitwise AND with the value 0x000000FF (which is called a "mask"), resulting of course in the same value 0x000000FF which represents 255.
If you think this is a nuisance then you are right. Java should have used unsigned bytes, but it decided to use only one particular integer type that is signed. Maybe a bit overzealous but still a big improvement over e.g. C where you have way too many integer formats, and each of them may be represented differently on various machines.
This question already has answers here:
Is a byte always 8 bits?
(3 answers)
Closed 3 years ago.
I understand byte is the underlying data type of Java IO but why byte is used to read and write as it has maximum value range -128 to 127. This range is integer, how can integers be used for reading and writing different symbols characters or binary data ?
Expectation is to understand why byte data type is used for Java IO.
Java was designed after C/C++ with some discussed topics:
Java text String (Reader/Writer) contains Unicode so text in mixed scripts may be combined. Internally String was an array of UTF-16 char; a .class file uses UTF-8 string constants. Hence byte[] (InputStream/OutputStream) is only for binary data. Between text and binary data there is always a conversion using the binary data's encoding/charset.
Numerical primitive types exist only in the signed version. (Except char that one could consider non-numeric.) The idea was to root out signed/unsigned "problems" of C++. So also byte is signed from -128 to 127. However overflow is irrelevant in java too,
so one can do:
byte b = (byte) 255; // 0xFF or -1
The primitive types byte/short/int/long have a fixed size of bytes, where C was notorious cross-platform, and things like C uint32 a bit ugly (32).
Having experienced tricky C bugs with signed/unsigned myself (before >10 years), I think this decision was okay.
It is easier to calculate in a signed mind set, and then at the end consider values as unsigned, than have throughout the expressions signed and unsigned parts.
Nowadays there is support in java for calculations respecting an unsigned interpretation of values, like Integer.parseUnsignedInt.
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hex to int number format exception in java
(4 answers)
Closed 6 years ago.
I get following exception:
java.lang.NumberFormatException: For input string: "3693019260"
while calling
Integer.parseInt(s);
And I don't know why I get it.
3693019260 is smaller than 2^32
3693019260 is clearly a natural number
the String is cleared from all non-digit elements with s.replaceAll("[^0-9]", "")
So why do I get this exception?
According to the little bit of debugging I did, I saw that the number dips under multmin but I don't know what this variable does and how I should interpret this observation.
While 3693019260 may fit into a 32 bit unsigned integer, it looks like you are trying to parse it into a plain int which is a signed integer. Signed simply means that it supports negative values using -.
With signed numbers, half of the namespace is reserved, so your number must fit into 2^32÷2−1 2147483647 instead of simply 2^32.
The simplest fix is to parse the value as a long instead of an int. Long numbers are 64 bits and support many more digits in the string.
This question already has answers here:
Assigning int to byte in java?
(5 answers)
Closed 8 years ago.
I have an int variable in java whose value is less than 256. So Can I store it in a Single byte variable.
I performed the following code.
int i =247;
byte b = (byte) i ;
But When I print it,
System.out.println(" i = "+(int)b);
It outputs
i = -9
Is it possible to convert int values less than 256 to single byte variables.
Pls. Help..
This isn't going to work. The MAX_VALUE for byte is 127 because bytes are signed in Java. (Consequently, the MIN_VALUE is -128. This is your typical Two's Compliment pattern.)
That being said, you have a couple options:
Leave it as an int and just use an int for your calculations. This might even be faster for you.
Leave it as a byte even if the number isn't correct. The numbers might not be correct in Java, but if all you care about is the bits being correct, they will be correct. So you can send that bite (which is a -9 in Java) across a webservice to another app that reads it as an unsigned byte, and it will read it as 247 correctly.
This question already has answers here:
What is the difference between Short and Character apart from processing?
(3 answers)
Closed 3 years ago.
According to the Java standard the short and char types both use 2 bytes so when I code something like:
char ch = 'c';
short s = ch;
There is an error saying "possible loss of precision". What am I missing here?
char is unsigned, short is signed.
So while they are both 2-byte long, they use the sixteenth bit for different purposes.
The range of the char type is 0 to 2^16 - 1 (0 to 65535).
The short range is -2^15 to 2^15 - 1 (−32,768 to 32,767).
The difference is that char is unsigned, short is signed. Thus, half the range of values of char is too big to be represented as a short (and of course, in symmetry, char cannot represent any of the negative values short can).