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Programming-Idioms

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  • Haskell

Idiom #77 Complex number

Declare a complex x and initialize it with value (3i - 2). Then multiply it by i.

The complex x is moved on the 2D plane when multiplied by i 
let i = ( 0 :+ 1 ) ; x = 3 * i - 2 in x * i
ímport Data.Complex
x = (0 :+ 1) * ((0 - 2) :+ 3)

Declaration and initialization is not required in Haskell.
As prefix negation in Haskell is tricky, I've used (0 - 2) for (-2).
Haskell variables are immutable, therefore "multiply it by two" is both unnecessary and too complicated.
with Ada.Numerics.Complex_Types;
declare
   use Ada.Numerics.Complex_Types;
   X : Complex := (Re => -2.0, Im => 3.0);
begin
   X := X * i;
end;
New implementation...
< > ⌨
deleplace 
let i = ( 0 :+ 1 ) ; x = 3 * i - 2 in x * i
x = (0 :+ 1) * ((0 - 2) :+ 3)
declare
   use Ada.Numerics.Complex_Types;
   X : Complex := (Re => -2.0, Im => 3.0);
begin
   X := X * i;
end;
float complex _x = -2 + 3 * I;

_x *= I;
using namespace std::complex_literals;
	
auto x = 3i - 2.;
x *= 1i;

var x = new Complex(-2, 3);
x *= Complex.ImaginaryOne;
auto x = complex(-2, 3);
x * complex(0, 1);
  complex :: x
  x = (-2,3)
  x = x * (0,1)

x := 3i - 2
x *= 1i
var x = math.complex(-2, 3);
x = math.multiply(x, math.i);
Complex x = new Complex(-2.0, 3.0);
x = x.multiply(Complex.I);
(let ((x #c(-2 3)))
  (* x #c(0 1)))
var
  C: Complex;
begin
  C := 3*i - 2;
end.
my $x = 3*i - 2;
$x *= i;
my $x = cplx(-2, 3);
$x *= i;
x = 3j-2
y = x * 1j
x = 3i - 2
x *= 1i
let mut x = Complex::new(-2, 3);
x *= Complex::i();
(define x -2+3i)
(display (* x 0+1i))