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Programming-Idioms

# 165 Last element of list
Assign to the variable x the last element of the list items.
Implementation
Ruby

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Other implementations 
x := items[len(items)-1]
const x = items[items.length - 1];
x = items[-1]
x = items[-1]
x := items[high(items)];
val x = items.last
foo :: [a] -> Maybe a
foo [] = Nothing
foo xs = Just $ last xs

x = foo items
int[42] items;
int x = items[$-1];
import std.range;
int[] items;
auto x = items.back();
x = items.last;
local x = items[#items]
let x = items[items.len()-1];
my $x = $items[-1];
$x = end($items);
let x = items.last().unwrap();
using System.Linq;
var x = items.LastOrDefault();
x = List.last(items)
int x = items[items.length - 1];
x = lists:last(items),
x = last items
(def x (last items))
#include<vector>
std::vector<int> items;
int last = items.back();
x = items(ubound(items,1))
var x = items.last()
$x = $items[array_key_last($items)];
#include <iterator>
auto x = *std::crbegin(items);
my $x = $items[$#items];
Dim x = items(items.Count - 1)
(def x (peek items))
var x = items[items.Count-1];
@import Foundation;
x=items.lastObject;
int length = sizeof(items) / sizeof(items[0]);
int x = items[length - 1];
X := Items'Last
(define items (list 1 2 3 4))
(define x (last items))
(setf x (car (last items)))
var x = items[^1];
const x = items.at(-1);
x := items last.
def x = items.last()

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