Declaring an Unsigned Int in Java

Declaring an unsigned int in Java

Java does not have a datatype for unsigned integers.

You can define a long instead of an int if you need to store large values.

You can also use a signed integer as if it were unsigned. The benefit of two's complement representation is that most operations (such as addition, subtraction, multiplication, and left shift) are identical on a binary level for signed and unsigned integers. A few operations (division, right shift, comparison, and casting), however, are different. As of Java SE 8, new methods in the Integer class allow you to fully use the int data type to perform unsigned arithmetic:

In Java SE 8 and later, you can use the int data type to represent an unsigned 32-bit integer, which has a minimum value of 0 and a maximum value of 2^32-1. Use the Integer class to use int data type as an unsigned integer. Static methods like compareUnsigned, divideUnsigned etc have been added to the Integer class to support the arithmetic operations for unsigned integers.

Note that int variables are still signed when declared but unsigned arithmetic is now possible by using those methods in the Integer class.

How to use the unsigned Integer in Java 8 and Java 9?

Per the documentation you posted, and this blog post - there's no difference when declaring the primitive between an unsigned int/long and a signed one. The "new support" is the addition of the static methods in the Integer and Long classes, e.g. Integer.divideUnsigned. If you're not using those methods, your "unsigned" long above 2^63-1 is just a plain old long with a negative value.

From a quick skim, it doesn't look like there's a way to declare integer constants in the range outside of +/- 2^31-1, or +/- 2^63-1 for longs. You would have to manually compute the negative value corresponding to your out-of-range positive value.

Why doesn't Java support unsigned ints?

This is from an interview with Gosling and others, about simplicity:

Gosling: For me as a language designer, which I don't really count myself as these days, what "simple" really ended up meaning was could I expect J. Random Developer to hold the spec in his head. That definition says that, for instance, Java isn't -- and in fact a lot of these languages end up with a lot of corner cases, things that nobody really understands. Quiz any C developer about unsigned, and pretty soon you discover that almost no C developers actually understand what goes on with unsigned, what unsigned arithmetic is. Things like that made C complex. The language part of Java is, I think, pretty simple. The libraries you have to look up.

Unsigned int (primitive) and Integer (Object) usage in Java

But when I tried this out, my compiler is complaining that the value I'm using is too large, 2^31.

You have an error because the literal 2147483648 is syntactically invalid.

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How exactly do I use an int/Integer to store unsigned value, such as 2^31?

You can still perform unsigned arithmetic using the new methods of the Integer class, just don't use invalid literals:

int n = 2147483647 + 1;  // 2^31

System.out.println(Integer.toUnsignedString(n));  // treat int as unsigned
System.out.println(Integer.toString(n));          // treat int as signed

2147483648
-2147483648

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What I'm understanding from this quote is that I can now store up to 2^32-1 as an int as long as I use the Integer wrapper object rather than the int primitive data type.

You can store the unsigned int using the primitive type int, but using the value as an unsigned type requires you to use the utility methods of Integer, as can be seen above.

How to output the absolute value of an Unsigned integer in java

You can view that integer as unsigned int by calling toUnsignedString():

int uint = Integer.parseUnsignedInt("4294967295");
System.out.println(Integer.toUnsignedString(uint));

You can also call some other methods that interpret that int as unsigned.

For example :

long l = Integer.toUnsignedLong(uint);
System.out.println(l); // will print 4294967295

int x = Integer.parseUnsignedInt("4294967295");
int y = 5;
int cmp1 = Integer.compareUnsigned(x,y); // interprets x as 4294967295 (x>y) 
int cmp2 = Integer.compare(x,y); // interprets x as -1 (x<y) 

Declaring an unsigned int in Java

Java does not have a datatype for unsigned integers.

You can define a long instead of an int if you need to store large values.

You can also use a signed integer as if it were unsigned. The benefit of two's complement representation is that most operations (such as addition, subtraction, multiplication, and left shift) are identical on a binary level for signed and unsigned integers. A few operations (division, right shift, comparison, and casting), however, are different. As of Java SE 8, new methods in the Integer class allow you to fully use the int data type to perform unsigned arithmetic:

In Java SE 8 and later, you can use the int data type to represent an unsigned 32-bit integer, which has a minimum value of 0 and a maximum value of 2^32-1. Use the Integer class to use int data type as an unsigned integer. Static methods like compareUnsigned, divideUnsigned etc have been added to the Integer class to support the arithmetic operations for unsigned integers.

Note that int variables are still signed when declared but unsigned arithmetic is now possible by using those methods in the Integer class.

Unsigned Int in Java

Depending on what you are doing, you can just treat long as a 64-bit value and int as a 32-bit value. Most operations esp readInt/Long writeInt/Long work just the same by ignoring the signness.

Can you give an example of operation you perform on these numbers and perhaps we can suggest how would do the same thing without having to expand the type.

For example, ++, --, +, -, *, ==, !=, << all work the same regardless of signess (i.e. give the same answer). for >> you can substitue >>>

It is the /, %, >, >=, <, <= and printing functions which assume signed values, but you should be able to work around these (if you use these).

e.g.

long unsignedA = 
long unsignedB = 
boolean greater = unsignedA + Long.MIN_VALUE > unsignedB + Long.MIN_VALUE

EDIT: Why does this work? Partly because java doesn't have overflow/underflow exceptions.

e.g.

byte unsignedA = 0;
unsignedA--; 
// unsignedA == FF, is this -1 or 255? Java assumes the former but you assume the later

byte unsignedB = unsignedA * unsignedA;
// unsignedB is -1 * -1 = 1 or (byte) (255*255) = (byte) 65525 = 1.

Unsigned short in Java

You can't, really. Java doesn't have any unsigned data types, except char.

Admittedly you could use char - it's a 16-bit unsigned type - but that would be horrible in my view, as char is clearly meant to be for text: when code uses char, I expect it to be using it for UTF-16 code units representing text that's interesting to the program, not arbitrary unsigned 16-bit integers with no relationship to text.

Unsigned int to byte

Noting that creating your own encryption is unsafe, and assuming you're doing it "just for fun" and don't think in any way that what you're doing is secure, you really don't need anything special...

int i = in.read();
byte b = (byte) i;
byte e = encrypt(b);
out.write(e);

would be the basic approach, assuming byte encrypt(byte b) method which does the "encryption". Checking for end-of-stream, exception handling, performance considerations (you don't want to perform things 1 byte at a time) etc. have been left out from this example.

`using my_type = unsigned;` vs `using my_type = unsigned int;` in C++

Yes they are identical. Using my_type can make the intention clearer. E.g. consider using celsius = int;

To get even better type safety you can use the newtype pattern (Google it) but it is a bit tedious in C++.

The choice of unsigned Vs unsigned int is just taste.

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