Idiom #256 Count backwards

Print the numbers 5, 4, ..., 0 (included), one line per number.

(dotimes (i 6) (print (- 5 i)))

for A in reverse 0 .. 5 loop
   Put_Line (A'Image);
end loop;

for (int i = 5; i >= 0; i--) {
	printf("%d\n", i);
}

(doseq [n (range 5 0 -1)]
  (println n))

for (int i = 5; i >= 0; --i) {
	std::cout << i;
}

using System;

for (int i = 5; i >= 0; i--)
{
    Console.WriteLine(i);
}

for (int i = 5; i >= 0; i--) {
  print(i);
}

do i=5,0,-1
  print *,i
end do

import "fmt"

for i := 5; i >= 0; i-- {
	fmt.Println(i)
}

import Control.Monad (forM_)

forM_ (reverse [0..5]) print

foldl  (\res x-> x:res) [] [0..5]

for (let i = 5; i >= 0; i--) {
  console.log(i)
}

for(int i=5 ; i>=0 ; i--) {
    System.out.println(i);
}

import static java.lang.System.out;
import static java.util.stream.IntStream.iterate;

iterate(5, i -> i - 1)
    .limit(6)
    .forEach(out::println);

import static java.lang.System.out;

for (int i = 5; i != -1; out.println(i--));

for i=5, 0, -1 do
	print(i)
end

$values = range(0, 5);
$valuesReversed = array_reverse($values);
foreach($valuesReversed as $value)
{
    printf("%s\n", $value);
}

for i := 5 downto 0 do writeln(i);

print "$_\n" for reverse (0..5);

print(*reversed(range(6)), sep='\n')

for i in range(5, -1, -1):
    print(i)

5.downto(0){|n| puts n }

for i in (0..=5).rev() {
    println!("{}", i);
}

(0..=5).rev().for_each(|i| println!("{}", i));

(let loop ([x 5])
  (when (>= x 0)
    (display x)
    (newline)
    (loop (sub1 x))))

5 to: 0 by: -1 do: [:i | Transcript showln: i asString].

Imports System

For i = 5 To 0 Step -1
    Console.WriteLine(i)
Next

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Idiom created by programming-idioms.org

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