How to make the mod of a negative number to be positive?

Question

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!

Answer 1

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

Answer 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.

Answer 3

Add the base if the input number X is negative:

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

Answer 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)

Answer 5

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.

Answer 6

int x=-3;

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

//second approach means just reverse the number modNum%x
cout<<5%x;