[Arnavion]

>This only works on two's complement machines.

No it does not. The behavior of -n where n is an unsigned integer is defined by the standard to be the result of subtracting n from 2^(number of bits in the type), independent of the characteristics of the machine.

This is true of both C and C++; and has been true since at least C99 and C++03

Random C++ standard draft from 2005: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n190... Section 5.3.1

>The operand of the unary - operator shall have arithmetic or enumeration type and the result is the negation of its operand. Integral promotion is performed on integral or enumeration operands. The negative of an unsigned quantity is computed by subtracting its value from 2^n, where n is the number of bits in the promoted operand. The type of the result is the type of the promoted operand.

Random C standard draft from 2007: http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf Section 6.2.5

>The range of nonnegative values of a signed integer type is a subrange of the corresponding unsigned integer type, and the representation of the same value in each type is the same. A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.

[gumby]

I don’t read those extracts as implying the same result as you do. The result type (promoted type in the C++ case) will still be unsigned, yet will be in most cases (especially small ones) full of 1s where there were zeros, and won’t compute the expected modulus. Draw out the bit patterns of 1’s vs 2’s complement for (say) n=5 and n=65535 and see if I have it wrong.

[Arnavion]

Not sure what you're trying to say. The point of what I wrote, and what the article says, is that `-n` is equivalent to `2^N - n`, and thus `-n % n` is equivalent to `(2^N - n) % n`.

If 2^N - n happens to be a number that's "full of 1s where there were zeros", then that's what it's supposed to be. I don't understand why you think it "won't compute the expected modulus". That's exactly what it will do.

If N is 32 and n is 5, then -n is 4294967291 and -n % n is 1. It doesn't matter whether that was calculated on a ones complement or twos complement machine.