Print the numbers 5, 4, ..., 0 (included), one line per number.
5.downto(0){|n| puts n }
for A in reverse 0 .. 5 loop Put_Line (A'Image); end loop;
for (int i = 5; i >= 0; i--) { printf("%d\n", i); }
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); }
(dotimes (i 6) (print (- 5 i)))
for i=5, 0, -1 do print(i) end
for i := 5 downto 0 do writeln(i);
print "$_\n" for reverse (0..5);
for i in range(5, -1, -1): print(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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