# Idiom #256 Count backwards

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

using System;
for (int i = 5; i >= 0; i--)
{
Console.WriteLine(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;
}
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)
}
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
\$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);
for i in range(5, -1, -1):
print(i)
5.downto(0){|n| puts n }
(0..=5).rev().for_each(|i| println!("{}", i));
for i in (0..=5).rev() {
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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