3 ms·

It is also worth noting that in most languages -2 / 3 yields 0 and not -1, i.e. integer devision rounds towards zero, not towards negative infinity.

• wyldfire 10y ago
  Yes, in fact that's the point I'm trying to focus on but never made explicit. Thanks.
  • tzs 10y ago
    Integer division rounding toward negative infinity makes it so that for integers a, b, with b > 0, then if a%b is defined as a - b * (a/b), you will always have 0 <= a%b < b. Integer division rounding toward zero makes that not so. a%b for negative a, positive b, will be <= 0. On the other hand, rounding toward negative infinity means that (-a)/b != -(a/b), when b does not divide a. Rounding toward zero makes (-a)/b = -(a/b). I think most mathematicians would rather have 0 <= a%b < b than (-a)/b = -(a/b) if they can't have both.
    • danbruc 10y ago
      Eric Lippert, a former C# compiler developer, has a good blog post [1] discussing the reasoning and the difference between modulus and remainder. [1] https://blogs.msdn.microsoft.com/ericlippert/2011/12/05/whats-the-difference-remainder-vs-modulus/ https://blogs.msdn.microsoft.com/ericlippert/2011/12/05/what...