Rounding integer division (instead of truncating)

问题

I was curious to know how I can round a number to the nearest whole number. For instance, if I had:

int a = 59 / 4;

which would be 14.75 if calculated in floating point; how can I store the result as 15 in \"a\"?

回答1:

int a = 59.0f / 4.0f + 0.5f;

This only works when assigning to an int as it discards anything after the '.'

Edit: This solution will only work in the simplest of cases. A more robust solution would be:

unsigned int round_closest(unsigned int dividend, unsigned int divisor)
{
    return (dividend + (divisor / 2)) / divisor;
}

回答2:

The standard idiom for integer rounding up is:

int a = (59 + (4 - 1)) / 4;

You add the divisor minus one to the dividend.

回答3:

A code that works for any sign in dividend and divisor:

int divRoundClosest(const int n, const int d)
{
  return ((n < 0) ^ (d < 0)) ? ((n - d/2)/d) : ((n + d/2)/d);
}

If you prefer a macro:

#define DIV_ROUND_CLOSEST(n, d) ((((n) < 0) ^ ((d) < 0)) ? (((n) - (d)/2)/(d)) : (((n) + (d)/2)/(d)))

The linux kernel macro DIV_ROUND_CLOSEST doesn't work for negative divisors!

回答4:

You should instead use something like this:

int a = (59 - 1)/ 4 + 1;

I assume that you are really trying to do something more general:

int divide(x, y)
{
   int a = (x -1)/y +1;

   return a;
}

x + (y-1) has the potential to overflow giving the incorrect result; whereas, x - 1 will only underflow if x = min_int...

回答5:

(Edited) Rounding integers with floating point is the easiest solution to this problem; however, depending on the problem set is may be possible. For example, in embedded systems the floating point solution may be too costly.

Doing this using integer math turns out to be kind of hard and a little unintuitive. The first posted solution worked okay for the the problem I had used it for but after characterizing the results over the range of integers it turned out to be very bad in general. Looking through several books on bit twiddling and embedded math return few results. A couple of notes. First, I only tested for positive integers, my work does not involve negative numerators or denominators. Second, and exhaustive test of 32 bit integers is computational prohibitive so I started with 8 bit integers and then mades sure that I got similar results with 16 bit integers.

I started with the 2 solutions that I had previously proposed:

#define DIVIDE_WITH_ROUND(N, D) (((N) == 0) ? 0:(((N * 10)/D) + 5)/10)

#define DIVIDE_WITH_ROUND(N, D) (N == 0) ? 0:(N - D/2)/D + 1;

My thought was that the first version would overflow with big numbers and the second underflow with small numbers. I did not take 2 things into consideration. 1.) the 2nd problem is actually recursive since to get the correct answer you have to properly round D/2. 2.) In the first case you often overflow and then underflow, the two canceling each other out. Here is an error plot of the two (incorrect) algorithms:

This plot shows that the first algorithm is only incorrect for small denominators (0 < d < 10). Unexpectedly it actually handles large numerators better than the 2nd version.

Here is a plot of the 2nd algorithm:

As expected it fails for small numerators but also fails for more large numerators than the 1st version.

Clearly this is the better starting point for a correct version:

#define DIVIDE_WITH_ROUND(N, D) (((N) == 0) ? 0:(((N * 10)/D) + 5)/10)

If your denominators is > 10 then this will work correctly.

A special case is needed for D == 1, simply return N. A special case is needed for D== 2, = N/2 + (N & 1) // Round up if odd.

D >= 3 also has problems once N gets big enough. It turns out that larger denominators only have problems with larger numerators. For 8 bit signed number the problem points are

if (D == 3) && (N > 75)) else if ((D == 4) && (N > 100)) else if ((D == 5) && (N > 125)) else if ((D == 6) && (N > 150)) else if ((D == 7) && (N > 175)) else if ((D == 8) && (N > 200)) else if ((D == 9) && (N > 225)) else if ((D == 10) && (N > 250))

(return D/N for these)

So in general the the pointe where a particular numerator gets bad is somewhere around
N > (MAX_INT - 5) * D/10

This is not exact but close. When working with 16 bit or bigger numbers the error < 1% if you just do a C divide (truncation) for these cases.

For 16 bit signed numbers the tests would be

if ((D == 3) && (N >= 9829)) else if ((D == 4) && (N >= 13106)) else if ((D == 5) && (N >= 16382)) else if ((D == 6) && (N >= 19658)) else if ((D == 7) && (N >= 22935)) else if ((D == 8) && (N >= 26211)) else if ((D == 9) && (N >= 29487)) else if ((D == 10) && (N >= 32763))

Of course for unsigned integers MAX_INT would be replaced with MAX_UINT. I am sure there is an exact formula for determining the largest N that will work for a particular D and number of bits but I don't have any more time to work on this problem...

(I seem to be missing this graph at the moment, I will edit and add later.) This is a graph of the 8 bit version with the special cases noted above:![8 bit signed with special cases for 0 < N <= 10 3

Note that for 8 bit the error is 10% or less for all errors in the graph, 16 bit is < 0.1%.

回答6:

As written, you're performing integer arithmetic, which automatically just truncates any decimal results. To perform floating point arithmetic, either change the constants to be floating point values:

int a = round(59.0 / 4);

Or cast them to a float or other floating point type:

int a = round((float)59 / 4);

Either way, you need to do the final rounding with the round() function in the math.h header, so be sure to #include <math.h> and use a C99 compatible compiler.

回答7:

From Linux kernel (GPLv2):

/*
 * Divide positive or negative dividend by positive divisor and round
 * to closest integer. Result is undefined for negative divisors and
 * for negative dividends if the divisor variable type is unsigned.
 */
#define DIV_ROUND_CLOSEST(x, divisor)(          \
{                           \
    typeof(x) __x = x;              \
    typeof(divisor) __d = divisor;          \
    (((typeof(x))-1) > 0 ||             \
     ((typeof(divisor))-1) > 0 || (__x) > 0) ?  \
        (((__x) + ((__d) / 2)) / (__d)) :   \
        (((__x) - ((__d) / 2)) / (__d));    \
}                           \
)

回答8:

int a, b;
int c = a / b;
if(a % b) { c++; }

Checking if there is a remainder allows you to manually roundup the quotient of integer division.

回答9:

#define CEIL(a, b) (((a) / (b)) + (((a) % (b)) > 0 ? 1 : 0))

Another useful MACROS (MUST HAVE):

#define MIN(a, b)  (((a) < (b)) ? (a) : (b))
#define MAX(a, b)  (((a) > (b)) ? (a) : (b))
#define ABS(a)     (((a) < 0) ? -(a) : (a))

回答10:

Here's my solution. I like it because I find it more readable and because it has no branching (neither ifs nor ternaries).

int32_t divide(int32_t a, int32_t b) {
  int32_t resultIsNegative = ((a ^ b) & 0x80000000) >> 31;
  int32_t sign = resultIsNegative*-2+1;
  return (a + (b / 2 * sign)) / b;
}

Full test program that illustrates intended behaviour:

#include <stdint.h>
#include <assert.h>

int32_t divide(int32_t a, int32_t b) {
  int32_t resultIsNegative = ((a ^ b) & 0x80000000) >> 31;
  int32_t sign = resultIsNegative*-2+1;
  return (a + (b / 2 * sign)) / b;
}

int main() {
  assert(divide(0, 3) == 0);

  assert(divide(1, 3) == 0);
  assert(divide(5, 3) == 2);

  assert(divide(-1, 3) == 0);
  assert(divide(-5, 3) == -2);

  assert(divide(1, -3) == 0);
  assert(divide(5, -3) == -2);

  assert(divide(-1, -3) == 0);
  assert(divide(-5, -3) == 2);
}

回答11:

Borrowing from @ericbn I prefere defines like

#define DIV_ROUND_INT(n,d) ((((n) < 0) ^ ((d) < 0)) ? (((n) - (d)/2)/(d)) : (((n) + (d)/2)/(d)))
or if you work only with unsigned ints
#define DIV_ROUND_UINT(n,d) ((((n) + (d)/2)/(d)))

回答12:

int divide(x,y){
 int quotient = x/y;
 int remainder = x%y;
 if(remainder==0)
  return quotient;
 int tempY = divide(y,2);
 if(remainder>=tempY)
  quotient++;
 return quotient;
}

eg 59/4 Quotient = 14, tempY = 2, remainder = 3, remainder >= tempY hence quotient = 15;

回答13:

double a=59.0/4;
int b=59/4;
if(a-b>=0.5){
    b++;
}
printf("%d",b);

let exact float value of 59.0/4 be x(here it is 14.750000)
let smallest integer less than x be y(here it is 14)
if x-y<0.5 then y is the solution
else y+1 is the solution

回答14:

try using math ceil function that makes rounding up. Math Ceil !

回答15:

If you're dividing positive integers you can shift it up, do the division and then check the bit to the right of the real b0. In other words, 100/8 is 12.5 but would return 12. If you do (100<<1)/8, you can check b0 and then round up after you shift the result back down.

回答16:

For some algorithms you need a consistent bias when 'nearest' is a tie.

// round-to-nearest with mid-value bias towards positive infinity
int div_nearest( int n, int d )
   {
   if (d<0) n*=-1, d*=-1;
   return (abs(n)+((d-(n<0?1:0))>>1))/d * ((n<0)?-1:+1);
   }

This works regardless of the sign of the numerator or denominator.

If you want to match the results of round(N/(double)D) (floating-point division and rounding), here are a few variations that all produce the same results:

int div_nearest( int n, int d )
   {
   int r=(n<0?-1:+1)*(abs(d)>>1); // eliminates a division
// int r=((n<0)^(d<0)?-1:+1)*(d/2); // basically the same as @ericbn
// int r=(n*d<0?-1:+1)*(d/2); // small variation from @ericbn
   return (n+r)/d;
   }

Note: The relative speed of (abs(d)>>1) vs. (d/2) is likely to be platform dependent.

回答17:

The following correctly rounds the quotient to the nearest integer for both positive and negative operands WITHOUT floating point or conditional branches (see assembly output below). Assumes N-bit 2's complement integers.

#define ASR(x) ((x) < 0 ? -1 : 0)  // Compiles into a (N-1)-bit arithmetic shift right
#define ROUNDING(x,y) ( (y)/2 - (ASR((x)^(y)) & (y)))

int RoundedQuotient(int x, int y)
   {
   return (x + ROUNDING(x,y)) / y ;
   }

The value of ROUNDING will have the same sign as the dividend (x) and half the magnitude of the divisor (y). Adding ROUNDING to the dividend thus increases its magnitude before the integer division truncates the resulting quotient. Here's the output of the gcc compiler with -O3 optimization for a 32-bit ARM Cortex-M4 processor:

RoundedQuotient:                // Input parameters: r0 = x, r1 = y
    eor     r2, r1, r0          // r2 = x^y
    and     r2, r1, r2, asr #31 // r2 = ASR(x^y) & y
    add     r3, r1, r1, lsr #31 // r3 = (y < 0) ? y + 1 : y
    rsb     r3, r2, r3, asr #1  // r3 = y/2 - (ASR(x^y) & y)
    add     r0, r0, r3          // r0 = x + (y/2 - (ASR(x^y) & y)
    sdiv    r0, r0, r1          // r0 = (x + ROUNDING(x,y)) / y
    bx      lr                  // Returns r0 = rounded quotient

回答18:

Some alternatives for division by 4

return x/4 + (x/2 % 2);
return x/4 + (x % 4 >= 2)

Or in general, division by any power of 2

return x/y + x/(y/2) % 2;    // or
return (x >> i) + ((x >> i - 1) & 1);  // with y = 2^i

It works by rounding up if the fractional part ⩾ 0.5, i.e. the first digit ⩾ base/2. In binary it's equivalent to adding the first fractional bit to the result

This method has an advantage in architectures with a flag register, because the carry flag will contain the last bit that was shifted out. For example on x86 it can be optimized into

shr eax, i
adc eax, 0

It's also easily extended to support signed integers. Notice that the expression for negative numbers is

(x - 1)/y + ((x - 1)/(y/2) & 1)

we can make it work for both positive and negative values with

int t = x + (x >> 31);
return (t >> i) + ((t >> i - 1) & 1);

回答19:

TLDR: Here's a macro; use it!

// To do (numer/denom), rounded to the nearest whole integer, use:
#define ROUND_DIVIDE(numer, denom) (((numer) + (denom) / 2) / (denom))

Usage example:

int num = ROUND_DIVIDE(13,7); // 13/7 = 1.857 --> rounds to 2, so num is 2

Full answer:

Some of these answers are crazy looking! Codeface nailed it though! (See @0xC0DEFACE's answer here). I really like the type-free macro or gcc statement expression form over the function form, however, so, I wrote this answer with a detailed explanation of what I'm doing (ie: why this mathematically works) and put it into 2 forms:

1. Macro form, with detailed commentary to explain the whole thing:

/// @brief      ROUND_DIVIDE(numerator/denominator): round to the nearest whole integer when doing 
///             *integer* division only
/// @details    This works on *integers only* since it assumes integer truncation will take place automatically
///             during the division! 
/// @notes      The concept is this: add 1/2 to any number to get it to round to the nearest whole integer
///             after integer trunction.
///             Examples:  2.74 + 0.5 = 3.24 --> 3 when truncated
///                        2.99 + 0.5 = 3.49 --> 3 when truncated
///                        2.50 + 0.5 = 3.00 --> 3 when truncated
///                        2.49 + 0.5 = 2.99 --> 2 when truncated
///                        2.00 + 0.5 = 2.50 --> 2 when truncated
///                        1.75 + 0.5 = 2.25 --> 2 when truncated
///             To add 1/2 in integer terms, you must do it *before* the division. This is achieved by 
///             adding 1/2*denominator, which is (denominator/2), to the numerator before the division.
///             ie: `rounded_division = (numer + denom/2)/denom`.
///             ==Proof==: 1/2 is the same as (denom/2)/denom. Therefore, (numer/denom) + 1/2 becomes 
///             (numer/denom) + (denom/2)/denom. They have a common denominator, so combine terms and you get:
///             (numer + denom/2)/denom, which is the answer above.
/// @param[in]  numerator   any integer type numerator; ex: uint8_t, uint16_t, uint32_t, int8_t, int16_t, int32_t, etc
/// @param[in]  denominator any integer type denominator; ex: uint8_t, uint16_t, uint32_t, int8_t, int16_t, int32_t, etc
/// @return     The result of the (numerator/denominator) division rounded to the nearest *whole integer*!
#define ROUND_DIVIDE(numerator, denominator) (((numerator) + (denominator) / 2) / (denominator))

2. GCC Statement Expression form:

See a little more on gcc statement expressions here.

/// @brief      *gcc statement expression* form of the above macro
#define ROUND_DIVIDE2(numerator, denominator)               \
({                                                          \
    __typeof__ (numerator) numerator_ = (numerator);        \
    __typeof__ (denominator) denominator_ = (denominator);  \
    numerator_ + (denominator_ / 2) / denominator_;         \
})