Idiom #165 Last element of list

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

foo :: [a] -> Maybe a
foo [] = Nothing
foo xs = Just $ last xs

x = foo items

x = last items

X := Items'Last

int length = sizeof(items) / sizeof(items[0]);
int x = items[length - 1];

(def x (peek items))

(def x (last items))

#include <iterator>

auto x = *std::crbegin(items);

#include<vector>

std::vector<int> items;
int last = items.back();

var x = items[items.Count-1];

var x = items[^1];

using System.Linq;

var x = items.LastOrDefault();

import std.range;

int[] items;
auto x = items.back();

int[42] items;
int x = items[$-1];

x = items.last;

x = List.last(items)

x = lists:last(items),

x = items(ubound(items,1))

x := items[len(items)-1]

def x = items.last()

const x = items[items.length - 1];

const x = items.at(-1);

T x = items.getLast();

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 = $items[$#items];

my $x = $items[-1];

x = items[-1]

x = items[-1]

x = items.last

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)

Do you know the best way to do this in your language ? New implementation...

< > ⌨

Idiom created by programming-idioms.org

History

Related idioms

Issues