Logo

Programming-Idioms

Assign to variable x the last element of list items.
New implementation

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating material.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations
X := Items'Last
int length = sizeof(items) / sizeof(items[0]);
int x = items[length - 1];
(def x (last items))
(def x (peek items))
#include <iterator>
auto x = *std::crbegin(items);
#include<vector>
std::vector<int> items;
int last = items.back();
var x = items[^1];
var x = items[items.Count-1];
using System.Linq;
var x = items.LastOrDefault();
int[42] items;
int x = items[$-1];
import std.range;
int[] items;
auto x = items.back();
x = items.last;
x = List.last(items)
x = lists:last(items),
x = items(ubound(items,1))
x := items[len(items)-1]
foo :: [a] -> Maybe a
foo [] = Nothing
foo xs = Just $ last xs

x = foo items
x = last items
const x = items.at(-1);
const x = items[items.length - 1];
int x = items[items.length - 1];
var x = items.last()
(setf x (car (last items)))
local x = items[#items]
@import Foundation;
x=items.lastObject;
$x = $items[array_key_last($items)];
$x = end($items);
x := items[high(items)];
my $x = $items[-1];
my $x = $items[$#items];
x = items[-1]
x = items.last
x = items[-1]
let x = items.last().unwrap();
let x = items[items.len()-1];
val x = items.last
(define items (list 1 2 3 4))
(define x (last items))
x := items last.
Dim x = items(items.Count - 1)