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Comment by _a_a_a_ | Hacker News Reader

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_a_a_a_

What 'mod' is for, innit.

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Right, cause why let the language do it automatically when you can use a bug prone manual implementation?

If you can't correctly implement a circular buffer with mod (or 'and') then the language can't save you, nothing can.

"If you can't correctly implement a for loop using assembly, then the language can't save you, nothing can"

Which is true. What's your point.

In c# -1 % 2 == -1. Doesn't help.

C#'s % is explicitly the remainder operator, not the modulo operator.

Not according to reflection.

    Expression<Func<int, int>> lambda = n => n % 2;
    Console.WriteLine(((dynamic)lambda).Body.NodeType);

Output is "Modulo".

Hmm... I haven't read the official spec, just the Microsoft documentation: https://learn.microsoft.com/en-us/dotnet/csharp/language-ref... In any case, the behavior is rem, not mod.

Ada has both rem and mod operatiors. I'm not sure how many other languages have operators for both.

Ecstasy uses % for modulo, and /% for divrem (division and remainder). So "a = b % c" for calculating the modulo, but "(a, r) = b /% c" to get the divisor and remainder.

It's a little verbose and can probably be reduced, but "((x % m) + m) % m" always works. Although it's probably not better than a check if x < 0 since branch prediction will get that right almost always.

I do think it's quite odd and frustrating that modulo can return negative numbers and I don't really get the reasoning there, but there's probably a good reason I don't know about.

The only reason is that C does it that way. And C does it that way because machine code does it that way.

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