Primitive type 'short' - casting in Java

Question

Hello,

I have a question about the primitive type 'short' in Java. I am using JDK 1.6.

If I have the following:

short a = 2;
short b = 3;
short c = a + b;

the compiler does not want to compile - it says that it "cannot convert from int to short" and suggests that I make a cast to short, so this:

short c = (short) (a + b);

really works. But my question is why do I need to cast? The values of a and b are in the range of short - the range of short values is {-32,768, 32767}. I also need to cast when I want to perform the operations -, *, / (I haven't checked for others).

If I do the same for primitive type int, I do not need to cast aa+bb to int. The following works fine:

int aa = 2;
int bb = 3;
int cc = aa +bb;

I discovered this while designing a class where I needed to add two variables of type short, and the compiler wanted me to make a cast. If I do this with two variables of type int, I don't need to cast.

Thank you very much in advance.

A small remark: the same thing also happens with the primitive type byte. So, this workes:

byte a = 2; byte b = 3; byte c = (byte) (a + b);

but this not:

byte a = 2; byte b = 3; byte c = a + b;

For long, float, double, and int, there is no need to cast. Only for short and byte values.

Answer 1

What compiler is this in? Works fine for me in MSVC++ 2008

Answer 2

What language are you using?

Many C based languages have a rule that any mathematical expression is performed in size int or larger. Because of this, once you add two shorts the result is of type int. This causes the need for a cast.

Answer 3

In C# and Java, the arithmatic expression on the right hand side of the assignment evaluates to int by default. That's why you need to cst back to a short, because there is no implicit conversion form int to short, for obvious reasons.

Answer 4

As explained in short C# (but also for other language compilers as well, like Java)

There is a predefined implicit conversion from short to int, long, float, double, or decimal.

You cannot implicitly convert nonliteral numeric types of larger storage size to short (see Integral Types Table for the storage sizes of integral types). Consider, for example, the following two short variables x and y:

short x = 5, y = 12;

The following assignment statement will produce a compilation error, because the arithmetic expression on the right-hand side of the assignment operator evaluates to int by default.

short z = x + y;   // Error: no conversion from int to short

To fix this problem, use a cast:

short z = (short)(x + y);   // OK: explicit conversion

It is possible though to use the following statements, where the destination variable has the same storage size or a larger storage size:

int m = x + y;
long n = x + y;

---

A good follow-up question is:

"why arithmetic expression on the right-hand side of the assignment operator evaluates to int by default" ?

A first answer can be found in:

Classifying and Formally Verifying Integer Constant Folding

The Java language specification defines exactly how integer numbers are represented and how integer arithmetic expressions are to be evaluated. This is an important property of Java as this programming language has been designed to be used in distributed applications on the Internet. A Java program is required to produce the same result independently of the target machine executing it.

In contrast, C (and the majority of widely-used imperative and object-oriented programming languages) is more sloppy and leaves many important characteristics open. The intention behind this inaccurate language specification is clear. The same C programs are supposed to run on a 16-bit, 32-bit, or even 64-bit architecture by instantiating the integer arithmetics of the source programs with the arithmetic operations built-in in the target processor. This leads to much more efficient code because it can use the available machine operations directly. As long as the integer computations deal only with numbers being “sufficiently small”, no inconsistencies will arise.

In this sense, the C integer arithmetic is a placeholder which is not defined exactly by the programming language specification but is only completely instantiated by determining the target machine.

Java precisely defines how integers are represented and how integer arithmetic is to be computed.

      Java Integers
--------------------------
Signed         |  Unsigned
--------------------------
long  (64-bit) |
int   (32-bit) |
short (16-bit) |  char (16-bit)
byte  (8-bit)  |

Char is the only unsigned integer type. Its values represent Unicode characters, from \u0000 to \uffff, i.e. from 0 to 2¹⁶−1.

If an integer operator has an operand of type long, then the other operand is also converted to type long. Otherwise the operation is performed on operands of type int, if necessary shorter operands are converted into int. The conversion rules are exactly specified.

[From Electronic Notes in Theoretical Computer Science 82 No. 2 (2003)
Blesner-Blech-COCV 2003: Sabine GLESNER, Jan Olaf BLECH,
Fakultät für Informatik,
Universität Karlsruhe
Karlsruhe, Germany]

Answer 5

Given that the "why int by default" question hasn't been answered ...

First, "default" is not really the right term (although close enough). As noted by VonC, an expression composed of ints and longs will have a long result. And an operation consisting of ints/logs and doubles will have a double result. The compiler promotes the terms of an expression to whatever type provides a greater range and/or precision in the result (floating point types are presumed to have greater range and precision than integral, although you do lose precision converting large longs to double).

One caveat is that this promotion happens only for the terms that need it. So in the following example, the subexpression 5/4 uses only integral values and is performed using integer math, even though the overall expression involves a double. The result isn't what you might expect...

(5/4) * 1000.0

OK, so why are byte and short promoted to int? Without any references to back me up, it's due to practicality: there are a limited number of bytecodes.

"Bytecode," as its name implies, uses a single byte to specify an operation. For example iadd, which adds two ints. Currently, 205 opcodes are defined, and integer math takes 18 for each type (ie, 36 total between integer and long), not counting conversion operators.

If short, and byte each got their own set of opcodes, you'd be at 241, limiting the ability of the JVM to expand. As I said, no references to back me up on this, but I suspect that Gosling et al said "how often do people actually use shorts?" On the other hand, promoting byte to int leads to this not-so-wonderful effect (the expected answer is 96, the actual is -16):

byte x = (byte)0xC0;
System.out.println(x >> 2);

Answer 6

I'd like to add something that hasn't been pointed out. Java doesn't take into account the values you have given the variables (2 and 3) in...

short a = 2; short b = 3; short c = a + b;

So as far as Java knows, you could done this...

short a = 32767; short b = 32767; short c = a + b;

Which would be outside the range of short, it autoboxes the result to an int becuase it's "possible" that the result will be more than a short but not more than an int. Int was chosen as a "default" because basically most people wont be hard coding values above 2,147,483,647 or below -2,147,483,648

Answer 7

In java, every numeric expression like:

anyPrimitive zas = 1;
anyPrimitive bar = 3;
?? x = zas  + bar 

x will always result to be at least an int, or a long if one of the addition elements was a long.

But there's are some quirks tough

byte a = 1; // 1 is an int, but it won't compile if you use a variable
a += 2; // the shortcut works even when 2 is an int
a++; // the post and pre increment operator work