Idiom #81 Round floating point number to integer

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

``````var y = x.round();
``````
``Y : constant Integer := Integer (Float'Rounding (X));``
``#include <math.h>``
``int y = (int)floorf(x + 0.5f);``
``(defn rnd [y] (int (+ y 0.5)))``
``#include <cmath>``
``int y = static_cast<int>(std::floor(x + 0.5f));``
``using System;``
``long y = (long)Math.Round(x);``
``import std.math: round;``
``int y = cast(int) x.round;``
``y = Kernel.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);``
``````(defun rnd (y) (floor (+ y 0.5)))
``````
``````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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