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Programming-Idioms

  • C
  • Elixir

Idiom #80 Truncate floating point number to integer

Declare integer y and initialize it with the value of floating point number x . Ignore non-integer digits of x .
Make sure to truncate towards zero: a negative x must yield the closest greater integer (not lesser).

y = trunc(x)
int y = (int)x;

The (int) isn't really necessary. The compiler will cast x automatically.
Y : constant Integer := Integer (Float'Truncation (X));
New implementation...
< > ⌨
deleplace 
y = trunc(x)
int y = (int)x;
Y : constant Integer := Integer (Float'Truncation (X));

int y = static_cast<int>(x);
int y = (int)x;
int y = cast(int)x;
int y = x.toInt()
integer :: y
real  :: x
y=x
y := int(x)
def y = x as int
y = truncate x

let y = BigInt (x | 0)
int y = (int)x;
(truncate x)

y = math.modf(x)
$y = (int)$x;
var
  y: integer;
  x: single = -0.8;
begin
  y := trunc(x);
end;
my $y = int($x);
y = int(x)
y = x.to_i
let y = x as i32;