Basic arithmetic operations on int - Java

Question

I recently noticed an idiosyncrasy of Java regarding basic arithmetic operations in Java. With the following code

byte a = 3;
byte b = 4;
byte c = a * b;

I get a "type mismatch" compilation error...

Are basic arithmetic operations in Java (+, -, *, /) only performed on primitive data types of int and higher order (long, double, etc.), whereas arithmetic operations on byte and short are first cast to int and then evaluated?

Good question. I honestly didn't know the answer, but found this. http://mindprod.com/jgloss/multiplication.html — Zutty, Jan 02 '13 at 16:53

Peter Lawrey · Accepted Answer · 2013-01-02T17:51:53.337

17

Operations on byte, char and short are widened to int unless the compiler can determine the value is in range.

final byte a = 3, b = 4;
byte c = a * b; // compiles

final byte a = 3, b = 40;
byte c = a * b; // compiles

final int a = 3, b = 4;
byte c = a * b; // compiles !!

but

byte a = 3, b = 4;
byte c = a * b; // doesn't compile as the result of this will be `int` at runtime.

final byte a = 30, b = 40;
byte c = a * b; // doesn't compile as the value is too large, will be an `int`

BTW This compiles even though it results in an overflow. :]

final int a = 300000, b = 400000;
int c = a * b; // compiles but overflows, is not made a `long`

edited Jan 02 '13 at 17:51

answered Jan 02 '13 at 16:53

Peter Lawrey

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Ouch, I didn't know `final` would make _such_ a difference here. But it looks strange anyway, since even in the second example, `a` and `b` are still declared as bytes... – fge Jan 02 '13 at 16:55
3

*unless the compiler can determine the value is in range* <=> it is a **constant** expression which is in range. – assylias Jan 02 '13 at 16:55

score 7 · Answer 2 · edited Jun 20 '20 at 09:12

The result of integer operations is either int or long. This is spelled out in the JLS:

4.2.2. Integer Operations

The numerical operators, which result in a value of type int or long:

The unary plus and minus operators + and - (§15.15.3, §15.15.4)

The multiplicative operators *, /, and % (§15.17)

The additive operators + and - (§15.18)

...

Also:

5.6.2. Binary Numeric Promotion

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order:

Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

If either operand is of type double, the other is converted to double.

Otherwise, if either operand is of type float, the other is converted to float.

Otherwise, if either operand is of type long, the other is converted to long.

Otherwise, both operands are converted to type int.

...

Binary numeric promotion is performed on the operands of certain operators:

The multiplicative operators *, / and % (§15.17)

The addition and subtraction operators for numeric types + and - (§15.18.2)

The numerical comparison operators <, <=, >, and >= (§15.20.1)

The numerical equality operators == and != (§15.21.1)

The integer bitwise operators &, ^, and | (§15.22.1)

In certain cases, the conditional operator ? : (§15.25)

Basic arithmetic operations on int - Java

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