C: Round signed integer to nearest multiple

Question

This feels like a basic question but I couldn't find a definitive answer so far.

I would like to implement an efficient function round_to_nearest(int x, int multiple), that rounds a signed integer x to the nearest multiple of multiple, avoiding using floating point arithmetic if possible.

Example output:

round_to_nearest(14, 5);
15

round_to_nearest(16, 5);
15

round_to_nearest(23, 5);
25

round_to_nearest(22, 5);
20

round_to_nearest(-23, 5);
-25

round_to_nearest(-22, 5);
-20

Answer 1

In integer arithmetic, if n is positive, add m/2, else subtract m/2, then divide by m (truncating integer divide), then multiply by m:

int round_to_nearest( int n, int m )
{
    return (( n + ((n < 0) ? -m : m) / 2) / m ) * m ;
}

int main()
{
    int test[] = {16, 23, 22, -23, -22} ;
    int m = 5 ;
    
    for( int i = 0; i < sizeof(test) / sizeof(*test); i++ )
    {
        printf(" round_to_nearest( %d, %d ) = %d\n", test[i], m, 
                                                     round_to_nearest( test[i], m ) ) ;
    }

    return 0;
}

Output of test:

 round_to_nearest( 16, 5 ) = 15
 round_to_nearest( 23, 5 ) = 25
 round_to_nearest( 22, 5 ) = 20
 round_to_nearest( -23, 5 ) = -25
 round_to_nearest( -22, 5 ) = -20

One caveat is that m must be > 0 - which in this context makes sense, I would accept that as a precondition for correct operation; checking for it as a runtime error is probably unnecessary, but you might include an assert to protect against programmer semantic error:

assert( m > 0 ) ;

Standard library asserts are removed when NDEBUG is defined - normally when debug support is disabled.

Answer 2

For positive numbers:

• add half of the multiple to x
• then perform integer division, which drops the fractional part
• then multiply by the multiple to get the final answer

For negative numbers, the first step is a subtraction, instead of addition.

int round_to_nearest(int x, int multiple)
{
    if (x >= 0)
        return ((x + multiple / 2) / multiple) * multiple;
    else
        return ((x - multiple / 2) / multiple) * multiple;
}