Mod of negative number is melting my brain

Question

I'm trying to mod an integer to get an array position so that it will loop round. Doing i % arrayLength works fine for positive numbers but for negative numbers it all goes wrong.

 4 % 3 == 1
 3 % 3 == 0
 2 % 3 == 2
 1 % 3 == 1
 0 % 3 == 0
-1 % 3 == -1
-2 % 3 == -2
-3 % 3 == 0
-4 % 3 == -1

so i need an implementation of

int GetArrayIndex(int i, int arrayLength)

such that

GetArrayIndex( 4, 3) == 1
GetArrayIndex( 3, 3) == 0
GetArrayIndex( 2, 3) == 2
GetArrayIndex( 1, 3) == 1
GetArrayIndex( 0, 3) == 0
GetArrayIndex(-1, 3) == 2
GetArrayIndex(-2, 3) == 1
GetArrayIndex(-3, 3) == 0
GetArrayIndex(-4, 3) == 2

I've done this before but for some reason it's melting my brain today :(

Answer 1

I always use my own mod function, defined as

int mod(int x, int m) {
    return (x%m + m)%m;
}

Of course, if you're bothered about having two calls to the modulus operation, you could write it as

int mod(int x, int m) {
    int r = x%m;
    return r<0 ? r+m : r;
}

or variants thereof.

The reason it works is that "x%m" is always in the range [-m+1, m-1]. So if at all it is negative, adding m to it will put it in the positive range without changing its value modulo m.

Answer 2

ShreevatsaR's answer won't work for all cases, even if you add "if(m<0) m=-m;", if you account for negative dividends/divisors.

For example, -12 mod -10 will be 8, and it should be -2.

The following implementation will work for both positive and negative dividends / divisors and complies with other implementations (namely, Java, Python, Ruby, Scala, Scheme, Javascript and Google's Calculator):

internal static class IntExtensions
{
    internal static int Mod(this int a, int n)
    {
        if (n == 0)
            throw new ArgumentOutOfRangeException("n", "(a mod 0) is undefined.");

        //puts a in the [-n+1, n-1] range using the remainder operator
        int remainder = a%n;

        //if the remainder is less than zero, add n to put it in the [0, n-1] range if n is positive
        //if the remainder is greater than zero, add n to put it in the [n-1, 0] range if n is negative
        if ((n > 0 && remainder < 0) ||
            (n < 0 && remainder > 0))
            return remainder + n;
        return remainder;
    }
}

---

Test suite using xUnit:

    [Theory]
    [PropertyData("GetTestData")]
    public void Mod_ReturnsCorrectModulo(int dividend, int divisor, int expectedMod)
    {
        Assert.Equal(expectedMod, dividend.Mod(divisor));
    }

    [Fact]
    public void Mod_ThrowsException_IfDivisorIsZero()
    {
        Assert.Throws<ArgumentOutOfRangeException>(() => 1.Mod(0));
    }

    public static IEnumerable<object[]> GetTestData
    {
        get
        {
            yield return new object[] {1, 1, 0};
            yield return new object[] {0, 1, 0};
            yield return new object[] {2, 10, 2};
            yield return new object[] {12, 10, 2};
            yield return new object[] {22, 10, 2};
            yield return new object[] {-2, 10, 8};
            yield return new object[] {-12, 10, 8};
            yield return new object[] {-22, 10, 8};
            yield return new object[] { 2, -10, -8 };
            yield return new object[] { 12, -10, -8 };
            yield return new object[] { 22, -10, -8 };
            yield return new object[] { -2, -10, -2 };
            yield return new object[] { -12, -10, -2 };
            yield return new object[] { -22, -10, -2 };
        }
    }

Answer 3

Single-line implementation using % only once:

int mod(int k, int n) {  return ((k %= n) < 0) ? k+n : k;  }

Answer 4

Please note that C# and C++'s % operator is actually NOT a modulo, it's remainder. The formula for modulo that you want, in your case, is:

float nfmod(float a,float b)
{
    return a - b * floor(a / b);
}

You have to recode this in C# (or C++) but this is the way you get modulo and not a remainder.

Answer 5

Comparing two predominant answers

(x%m + m)%m;

and

int r = x%m;
return r<0 ? r+m : r;

Nobody actually mentioned the fact that the first one may throw an OverflowException while the second one won't. Even worse, with default unchecked context, the first answer may return the wrong answer (see mod(int.MaxValue - 1, int.MaxValue) for example). So the second answer not only seems to be faster, but also more correct.

Answer 6

I like the trick presented by Peter N Lewis on this thread: "If n has a limited range, then you can get the result you want simply by adding a known constant multiple of [the divisor] that is greater that the absolute value of the minimum."

So if I have a value d that is in degrees and I want to take

d % 180f

and I want to avoid the problems if d is negative, then instead I just do this:

(d + 720f) % 180f

This assumes that although d may be negative, it is known that it will never be more negative than -720.