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Idiom #19 Reverse a list

Reverse the order of the elements of the list x.
This may reverse "in-place" and destroy the original ordering.

import std.range;
auto y = x.retro;

This doesn't allocate a new list or modify the original, but creates an iterator that traverses the list in reverse order.
import std.algorithm;
reverse(x);

Reverses x in-place.
with Ada.Containers.Vectors;
use Ada.Containers;
X.Reverse_Elements;
New implementation...
< > ⌨
programming-idioms.org 
auto y = x.retro;
reverse(x);
X.Reverse_Elements;
int *p1 = x;
int *p2 = x + N-1;

while (p1 < p2)
{
    int temp = *p1;
    *(p1++) = *p2;
    *(p2--) = temp;
}
List.rev x
(reverse x)
std::reverse(begin(x), end(x));
x.Reverse();
x = x.reversed.toList();
x = Enum.reverse(x)
lists:reverse(List)
  a = a(ubound(a,dim=1)::-1)
slices.Reverse(x)
for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
	x[i], x[j] = x[j], x[i]
}
func reverse[T any](x []T) {
	for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
		x[i], x[j] = x[j], x[i]
	}
}
x = x.reverse()
reverse x

x = x.reverse();
Collections.reverse(x);
Collections.reverse(x);
int i, m = x.size(), n = m-- / 2;
for (i = 0; i < n; ++i) swap(x, i, m - i);
reverse(x);
static <T> void reverse(List<T> x){
	int n = x.size();
	for(int i=0;i<n/2;i++){
		T tmp = x.get(i);
		x.set(i, x.get(n-i-1));
		x.set(n-i-1, tmp);
	}
}
x = x.reversed()
x.reverse()
val reversedView = x.asReversed()
(reverse x)
rev = {}
for i=#x, 1, -1 do
	rev[#rev+1] = x[i]
end

-- in-situ reversal
function reverse(t)
  local n = #t
  local i = 1
  for i = 1, n do
    t[i],t[n] = t[n],t[i]

    n = n - 1
  end
end
function array_reverse(x)
  local n, m = #x, #x/2
  for i=1, m do
    x[i], x[n-i+1] = x[n-i+1], x[i]
  end
  return x
end

rev = {}
for i=#x, 1, -1 do
	rev[#rev+1] = x[i]
end
x = rev
x.reverseObjectEnumerator.allObjects
$x = array_reverse($x, true);
var n,i: integer;
[...]
n:= length(x);
setlength(x, n+1);
for i := 0 to (n div 2)-1 do
	begin
		x[n] := x[i];
		x[i] := x[n-i-1];
		x[n-i-1] := x[n];
	end;
setlength(x, n);
var n,i: integer;
	tmp:TElement;
[...]
n:= length(x);
for i := 0 to (n div 2)-1 do
	begin
		tmp := x[i];
		x[i] := x[n-i-1];
		x[n-i-1] := tmp;
	end;
my @list = ('words', 'of', 'list', 'a', 'reverse');
my @reversed = reverse @list;
reverse(x, Reversed).
x.reverse()
x = reversed(x)
y = x[::-1]
x.reverse!
x.reverse();
let y: Vec<_> = x.into_iter().rev().collect();
x = x.reverse
(reverse x)
X reversed.
Dim ItemList As New List(Of String)(New String() {"one", "two", "three"})
ItemList.Reverse()
For Each item In ItemList
    Console.WriteLine(item)
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