Creating implementation for idiom 165 : Last element of list

Idiom

# 165 Last element of list

Created by

Lead paragraph

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

New implementation

Language

Imports

Code

Documentation URL

Original attribution URL

Online runnable demo

Username

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 (peek items))

(def x (last items))

#include <iterator>

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

#include<vector>

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

using System.Linq;

var x = items.LastOrDefault();

var x = items[^1];

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

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]

def x = items.last()

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

x = foo items

x = last items

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

const x = items.at(-1);

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

T x = items.getLast();

var x = items.last()

(setf x (car (last items)))

local x = items[#items]

@import Foundation;

x=items.lastObject;

$x = end($items);

$x = $items[array_key_last($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[items.len()-1];

let x = items.last().unwrap();

val x = items.last

(define items (list 1 2 3 4))
(define x (last items))

x := items last.

Dim x = items(items.Count - 1)