Programming-Idioms

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Idiom #81 Round floating point number to integer

Declare integer y and initialize it with the rounded value of floating point number x .
Ties (when the fractional part of x is exactly .5) must be rounded up (to positive infinity).

#include <math.h>
int y = (int)floorf(x + 0.5f);
#include <cmath>
int y = static_cast<int>(std::floor(x + 0.5f));
int x = Math.Round(y);
import std.math: round;
int y = cast(int) x.round;
var y = x.round();

y = Float.round x
integer :: y

y = nint(x)
import "math"
y := int(math.Floor(x + 0.5))
y = floor (x + 1/2)

var y = Math.round(x);
long y = Math.round(x);
function round(float)
    local int, part = math.modf(float)
    if float == math.abs(float) and part >= .5 then return int+1    -- positive float
    elseif part <= -.5 then return int-1                            -- negative float
    end
    return int
end
function round(float)
    return math.floor(float + .5)
end
$y = (int) round($x);
var
  y: integer;
  x: double;
begin
  y := round(x);
end.
my $y = int($x + 1/2);
y = int(x + 0.5)
y = (x + 1/2r).floor
let y = x.round() as i64;
(define y (round x))
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Idiom created by

deleplace

Related idioms

int y = (int)floorf(x + 0.5f);
int y = static_cast<int>(std::floor(x + 0.5f));
int x = Math.Round(y);
int y = cast(int) x.round;
var y = x.round();

y = Float.round x
integer :: y

y = nint(x)
y := int(math.Floor(x + 0.5))
y = floor (x + 1/2)

var y = Math.round(x);
long y = Math.round(x);
function round(float)
    local int, part = math.modf(float)
    if float == math.abs(float) and part >= .5 then return int+1    -- positive float
    elseif part <= -.5 then return int-1                            -- negative float
    end
    return int
end
function round(float)
    return math.floor(float + .5)
end
$y = (int) round($x);
var
  y: integer;
  x: double;
begin
  y := round(x);
end.
my $y = int($x + 1/2);
y = int(x + 0.5)
y = (x + 1/2r).floor
let y = x.round() as i64;
(define y (round x))