Question

This loop will continue indefinitely:

char a = 100;
for(a=100; a>=0;--a)
    System.out.println(a);

Does it happen because a gets promoted to an int value for arithmetic operations and gets widened to 32 bits from 16 bit char value and hence will always stay positive?

Answer 1

It will indeed loop indefinitely — and the reason you stated is close. It happens because a can’t represent any number that doesn’t satisfy a >= 0 — char is unsigned. Arithmetic underflow is well-defined in Java and unindicated. See the below relevant parts of the specification.

• §4.2.2

  The integer operators do not indicate overflow or underflow in any way.

  This means there is no indication of overflow/underflow other than just comparing the values… e.g. if a <= --a, then that means an underflow occured.

• §15.15.2

  Before the subtraction, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the difference is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored. The value of the prefix decrement expression is the value of the variable after the new value is stored.

  So, we can see that there are two major steps here: binary numeric promotion, followed by a narrowing pritimive conversion.

• §5.6.2

  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.

  We can see that the decrement expression works with a treated as int, thus performing a widening conversion. This allows it to represent the value -1.

• §5.1.3

  A narrowing primitive conversion may lose information about the overall magnitude of a numeric value and may also lose precision and range.

  …

  A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the sign of the resulting value to differ from the sign of the input value.

  Keeping only the n lowest order bits means that only the lowest 16 bits of the int expression a - 1 are kept. Since -1 is 0b11111111 11111111 11111111 11111111 here, only the lower 0b11111111 11111111 is saved. Since char is unsigned, all of these bits contribute to the result, giving 65535.

Noticing something here? Essentially, this all means that Java integer arithmetic is modular; in this case, the modulus is 2^16, or 65536, because char is a 16-bit datatype. -1 (mod 65536) ≡ 65535, so the decrement will wrap back around.