converting -1 to unsigned types

Question

Consider the following code to set all bits of x

unsigned int x = -1;

Is this portable ? It seems to work on at least Visual Studio 2005-2010

Answer 1

The citation-heavy answer:

I know there are plenty of correct answers in here, but I'd like to add a few citations to the mix. I'll cite two standards: C99 n1256 draft (freely available) and C++ n1905 draft (also freely available). There's nothing special about these particular standards, they're just both freely available and whatever happened to be easiest to find at the moment.

The C++ version:

§5.3.2 ¶9: According to this paragraph, the value ~(type)0 is guaranteed to have all bits set, if (type) is an unsigned type.

The operand of ~ shall have integral or enumeration type; the result is the one’s complement of its operand. Integral promotions are performed. The type of the result is the type of the promoted operand.

§3.9.1 ¶4: This explains how overflow works with unsigned numbers.

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2ⁿ where n is the number of bits in the value representation of that particular size of integer.

§3.9.1 ¶7, plus footnote 49: This explains that numbers must be binary. From this, we can infer that ~(type)0 must be the largest number representable in type (since it has all bits turned on, and each bit is additive).

The representations of integral types shall define values by use of a pure binary numeration system⁴⁹.

49) A positional representation for integers that uses the binary digits 0 and 1, in which the values represented by successive bits are additive, begin with 1, and are multiplied by successive integral power of 2, except perhaps for the bit with the highest position. (Adapted from the American National Dictionary for Information Processing Systems.)

Since arithmetic is done modulo 2ⁿ, it is guaranteed that (type)-1 is the largest value representable in that type. It is also guaranteed that ~(type)0 is the largest value representable in that type. They must therefore be equal.

The C99 version:

The C99 version spells it out in a much more compact, explicit way.

§6.5.3 ¶3:

The result of the ~ operator is the bitwise complement of its (promoted) operand (that is, each bit in the result is set if and only if the corresponding bit in the converted operand is not set). The integer promotions are performed on the operand, and the result has the promoted type. If the promoted type is an unsigned type, the expression ~E is equivalent to the maximum value representable in that type minus E.

As in C++, unsigned arithmetic is guaranteed to be modular (I think I've done enough digging through standards for now), so the C99 standard definitely guarantees that ~(type)0 == (type)-1, and we know from §6.5.3 ¶3 that ~(type)0 must have all bits set.

The summary:

Yes, it is portable. unsigned type x = -1; is guaranteed to have all bits set according to the standard.

Footnote: Yes, we are talking about value bits and not padding bits. I doubt that you need to set padding bits to one, however. You can see from a recent Stack Overflow question (link) that GCC was ported to the PDP-10 where the long long type has a single padding bit. On such a system, unsigned long long x = -1; may not set that padding bit to 1. However, you would only be able to discover this if you used pointer casts, which isn't usually portable anyway.