Using Remainder Operator with Negative Divident

[valentin tihomirov]

My rateonale is:
-1 mod 3 = must be 3
0 mod 3 = 0
1 mod 3 = 1
2 mod 3 = 2
3 mod 3 = 3
4 mod 3 = 0
= 1
= 2
= 3
= 0

We note a sequence: 0 1 2 3 0 1 2 3 0 1 2 3. Going in backward direction, we
get 3 2 1 0 3 2 1 0 3 2 1 0.

Since (0 % N) is 0, going one step back (-1 % N) must result in N. But c#
compiler gives me -1. It is wrong. Fix it. The C#LS seems wrong. Correct it.

[Jon Skeet [C# MVP]]

Re: Using Remainder Operator with Negative Divident

valentin tihomirov <V_tihomirov@be st.eewrote:

My rateonale is:
-1 mod 3 = must be 3
0 mod 3 = 0
1 mod 3 = 1
2 mod 3 = 2
3 mod 3 = 3
4 mod 3 = 0
= 1
= 2
= 3
= 0
>
We note a sequence: 0 1 2 3 0 1 2 3 0 1 2 3. Going in backward direction, we
get 3 2 1 0 3 2 1 0 3 2 1 0.
>
Since (0 % N) is 0, going one step back (-1 % N) must result in N. But c#
compiler gives me -1. It is wrong. Fix it. The C#LS seems wrong. Correct it.

Your understanding of "remainder" is too rigid. Fix it :)

Seriously, there are different understandings of how remainders should
work with negative numbers. Some systems use the sign of the dividend,
others use the sign of the divisor. See
http://en.wikipedia.org/wiki/Modulo_operation for a list.

You know what I think is *really* broken? To leave it to the
implementation - which is what C++ did, at least at the time of
Stroustrup's 2nd edition. (I seriously hope it's been fixed since
then.)


(I have in my head that "remainder" should work as shown in C#, but if
a language defines a "modulo" operator, that should always be non-
negative. However, I can't find any evidence backing that up.)

--
Jon Skeet - <skeet@pobox.co m>
http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet
If replying to the group, please do not mail me too

[Doug Semler]

Re: Using Remainder Operator with Negative Divident

"Jon Skeet [C# MVP]" <skeet@pobox.co mwrote in message
news:MPG.2160f0 a1540996f14b0@m snews.microsoft .com...

>
>
(I have in my head that "remainder" should work as shown in C#, but if
a language defines a "modulo" operator, that should always be non-
negative. However, I can't find any evidence backing that up.)

What should be non-negative? The result of the modulus operator? If the
dividend is negative in C#, the result of the modulus is negative, which
actually makes sense mathematically (what would you need to ADD to the
result of the divisor times the result of the division to get back to your
dividend?) Note this is only for integer modulus. I always hated float
modulus :)

-3 / -2 = 1 - what would you need to add? (-1) so -3 % -2 = -1
-3 / 2 = -1 - what do you add? -1 again. so -3 % 2 = -1
3 / -2 = -1 - what do you need to add to -1 * -2?


--
Doug Semler, MCPD
a.a. #705, BAAWA. EAC Guardian of the Horn of the IPU (pbuhh).
The answer is 42; DNRC o-
Gur Hfrarg unf orpbzr fb shyy bs penc gurfr qnlf, abbar rira
erpbtavmrf fvzcyr guvatf yvxr ebg13 nalzber. Fnq, vfa'g vg?

[Doug Semler]

Re: Using Remainder Operator with Negative Divident

"Doug Semler" <dougsemler@gma il.comwrote in message
news:evaBB9i$HH A.4612@TK2MSFTN GP03.phx.gbl...

"Jon Skeet [C# MVP]" <skeet@pobox.co mwrote in message
news:MPG.2160f0 a1540996f14b0@m snews.microsoft .com...
>

>>
>>
>(I have in my head that "remainder" should work as shown in C#, but if
>a language defines a "modulo" operator, that should always be non-
>negative. However, I can't find any evidence backing that up.)

>
>
What should be non-negative? The result of the modulus operator? If the
dividend is negative in C#, the result of the modulus is negative, which
actually makes sense mathematically (what would you need to ADD to the
result of the divisor times the result of the division to get back to your
dividend?) Note this is only for integer modulus. I always hated float
modulus :)

Note that my note above was for the example below, not that float modulus
acted differently with respect to sign <g>

>
-3 / -2 = 1 - what would you need to add? (-1) so -3 % -2 = -1
-3 / 2 = -1 - what do you add? -1 again. so -3 % 2 = -1
3 / -2 = -1 - what do you need to add to -1 * -2?
>
>

--
Doug Semler, MCPD
a.a. #705, BAAWA. EAC Guardian of the Horn of the IPU (pbuhh).
The answer is 42; DNRC o-
Gur Hfrarg unf orpbzr fb shyy bs penc gurfr qnlf, abbar rira
erpbtavmrf fvzcyr guvatf yvxr ebg13 nalzber. Fnq, vfa'g vg?

[Jon Skeet [C# MVP]]

Re: Using Remainder Operator with Negative Divident

Doug Semler <dougsemler@gma il.comwrote:

(I have in my head that "remainder" should work as shown in C#, but if
a language defines a "modulo" operator, that should always be non-
negative. However, I can't find any evidence backing that up.)

>
What should be non-negative? The result of the modulus operator?

Yes - in my head, anyway :)

If the dividend is negative in C#, the result of the modulus is negative, which
actually makes sense mathematically (what would you need to ADD to the
result of the divisor times the result of the division to get back to your
dividend?) Note this is only for integer modulus. I always hated float
modulus :)

Yes, I totally see your point. Of course, it depends on how you round
the division aspect. If you define integer division to round down, i.e.
to always return the largest number d such that d*x < y, then you will
always have a non-negative integer r such that d*x+r = y. However, C#
(and many other computer languages) use round-towards-zero for
division, which is where the negative remainder comes in.

At least, that's what I currently think - but I'm about to go to bed,
so anything I write now probably shouldn't be trusted.

--
Jon Skeet - <skeet@pobox.co m>
http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet
If replying to the group, please do not mail me too

[Jon Skeet [C# MVP]]

Re: Using Remainder Operator with Negative Divident

On Sep 24, 9:27 am, "valentin tihomirov" <V_tihomi...@be st.eewrote:

Actually, I have determined that their algorithm can be implemented without
wrapping the last column to 0 forth and back, which require these ugly
if-else. It is the Royal Mail User Guide, which must be fixed then
(simplified).

I'd prefer their algorithm to be fixed - keep the content of the table
where it is, but make 0 the first row/column. Too late to do that now,
but clearly you can remap the table if you're willing to change where
the contents are. It's only a lookup, after all.

Jon

[valentin tihomirov]

Re: Using Remainder Operator with Negative Divident

I'd prefer their algorithm to be fixed - keep the content of the table
where it is, but make 0 the first row/column. Too late to do that now,
but clearly you can remap the table if you're willing to change where
the contents are. It's only a lookup, after all.

The algorithm seems correctly designed. The explanation is bad moving the 0
column into the end and back for no reason.

[Jon Skeet [C# MVP]]

Re: Using Remainder Operator with Negative Divident

On Sep 24, 10:21 am, "valentin tihomirov" <V_tihomi...@be st.eewrote:

I'd prefer their algorithm to be fixed - keep the content of the table
where it is, but make 0 the first row/column. Too late to do that now,
but clearly you can remap the table if you're willing to change where
the contents are. It's only a lookup, after all.

>
The algorithm seems correctly designed. The explanation is bad moving the 0
column into the end and back for no reason.

Yes, by "algorithm" I guess I really meant "data used by the
algorithm" - i.e. the table.

Either way, it's certainly not C#'s fault :)

Jon

[james.curran@gmail.com]

Re: Using Remainder Operator with Negative Divident

On Sep 23, 5:53 pm, Jon Skeet [C# MVP] <sk...@pobox.co mwrote:

You know what I think is *really* broken? To leave it to the
implementation - which is what C++ did, at least at the time of
Stroustrup's 2nd edition. (I seriously hope it's been fixed since
then.)

I doubt it. Especially considering the C++ standard still does not
require a particular binary representation for -1, or, for that
matter, a particular value for 32767+1.