27

Basically, I need (-3) % 5 to be "2" instead of "-3". Python produces "2", but C++ produces "-3". Not sure how to produce "2" in C++. Thanks!

David G
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Hailiang Zhang
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  • Keith raises an important point in a comment. Is your second operand always positive? If not, what should happen when it's negative? – ysth Dec 10 '12 at 03:11

6 Answers6

33

Most easily: ((x % 5) + 5) % 5

ysth
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9

Add the base if the input number X is negative:

X % Y + (X % Y < 0 ? Y : 0);
perreal
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6

The quick & dirty way is to write

((x % divisor) + divisor) % divisor

For example, ((-3 % 5) + 5) % 5 == 2. However this performs two separate divisions, and since divisions are one of the slowest arithmetic operations you might like one of these alternatives:

(1) General mod for integer or floating point

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

template<class Num> Num mod(Num x, Num divisor)
{
    Num m = x % divisor;
    return m + (m < 0 ? divisor : 0);
}

(2) Non-branching mod for 32-bit integers

int mod(int x, int divisor)
{
    int m = x % divisor;
    return m + ((m >> 31) & divisor);
}

All this assumes that the divisor is always positive.

Qwertie
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4

You can add some multiple of 5 to the negative number first, to convert it to a positive number with the same value mod 5.

You can do that by taking the absolute of the negative number, adding whatever is needed to round it up to the next multiple of 5, and then add it to your negative number, and that should already be a number between 0 and 4.

Alternatively, simply do something like:

num = -2;
mod = 5;
if ( num < 0 ) {
    result = mod - (abs(num) % mod);
}

and it'll work (explanation: mathemagic)

sampson-chen
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  • @djechlin: I don't think you followed what sampson-chen was suggesting, something like: `(x + (abs(x)+y-abs(x)%y)) % y` – ysth Dec 10 '12 at 02:36
  • @djechlin the point of the absolute part is just so you start with some number bigger in magnitude than the negative number, and then you move up to the closest multiple of 5. That actually does preserve the value mod 5. Sorry if my wording was confusing in the original answer =p – sampson-chen Dec 10 '12 at 02:37
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    This should be the correct answer, since it's the only one that works if it's off by more than 1 multiple of `mod`. As Hailiang Zhang said, the code would fail if num%mod==0; I fixed it in an edit just now. – Alex Meiburg Jul 18 '19 at 18:01
  • @AlexMeiburg no idea what you are talking about. Maybe comment on my answer if you feel it is incorrect, with an example? – ysth Jan 26 '20 at 07:32
0
int x=-3;

// first approach
cout<<((x % 5) + 5) % 5;

//second approach means just reverse the number modNum%x
cout<<5%x;
shubham kapoor
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-2

I see a lot of suggestions for ((x % 5) + 5) % 5 But I'm getting the same result with just (X + 5) % 5

(X + divisor) % divisor.

Sidney CQ
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    This will not work if `X + divisor < 0`, for example, for `-13 % 5`. – zkoza Feb 08 '21 at 17:44
  • @zkoza you're right, it only works for me because I'm never adding or subtracting numbers smaller than my divisor. Good point. – Sidney CQ Feb 09 '21 at 18:15