Chad Nelson wrote:
On 04/02/2010 02:32 PM, Jeffrey Hellrung wrote:
Don't forget (assuming a single unsigned 0):
x / 0 => NaN
Eh? I thought we'd decided that anything divided by zero, other than zero itself, was infinity (or negative infinity, if x is negative)?
Of course you can define whatever arithmetic you like, but I think it makes more sense for products and quotients of like signs being positive, of opposite signs being negative, and anything involving (unsigned) zero giving either (unsigned) zero or NaN. [...]
Another topic: Chad, how natural is it, implementation-wise, to have a signed zero? I'm guessing you have a sign bit, leading naturally to either providing both a +0 or a -0, or ensuring after each operation that -0 is normalized to +0.
It wouldn't be a problem to implement it, and yes, I do have a sign bit (or rather, a sign Boolean, but that's essentially the same thing).
What are the arguments against a signed zero?
That there's no need for it in an integer library. [...] As such, there's no real need for it in XInt.
I was looking more for objections based purely on logical inconsistencies. Can you think of any?
Is it possible to simply be agnostic to the sign bit when the mantissa is 0?
Isn't that the same as saying that -0 is identical to +0?
If you don't have any infinities or NaNs to worry about, yes! For this case, I'm just suggesting you may not have to normalize zeros if -0 behaves exactly the same as +0. But it still may be useful to distinguish between them for libraries built on top of the core implementation. An arbitrary-precision floating point type may make a distinction between +0 and -0, and it would be convenient if the underlying integer core allowed this distinction naturally. Additionally, in the presence of an infinity value, you can then say x / +0 = +infinity x / -0 = -infinity if x > 0. Other than that (and maybe a few more exceptional cases), +0 and -0 behave identically. This has the (minute) advantage that you have simpler sign relationships on divisions: + / + = - / - = + + / - = - / + = - NaN / x = x / NaN = NaN This really only makes sense if your natural representations for +0 and -0 naturally behave identically, thus incurring no overhead (and perhaps simplifying implementation by avoiding a renormalization to "+0"). What do you think? - Jeff