Re: -1%N = -1

OuFeRRaT wrote:

> Does anyone know why the remainder and the division operations are
> incorrect with negative numbers in C++? Is it about assembly?

They aren't incorrect. They're just not required to follow the
definition of "modulus", which is why "%" is called the remainder
operator instead.  In fact, it's implementation defined. However, C99
(IIRC) defines the result of a%b to have the sign of a, therefore I'd
expect this to sneak into most C++ implementations as well.

This is probably even reasonable. The basic equality I'd expect to hold
is a==b*(a/b)+(a%b). If the sign of the remainder would follow the sign
of the divisor, it would be necessary that -10/3==-4 and -10%3==2. In
other words, ((a!=0) == ((a/b) != -(-a/b))) would always be true, quite
uncomfortably.

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