Logo

Programming-Idioms

This language bar is your friend. Select your favorite languages!
  • Scheme

Idiom #165 Last element of list

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

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

This runs in O(n) time since list produces a linked list.
X := Items'Last

X must be declared earlier.
New implementation...
< > ⌨
programming-idioms.org 
(define items (list 1 2 3 4))
(define 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))
auto x = *std::crbegin(items);
std::vector<int> items;
int last = items.back();
var x = items.LastOrDefault();
var x = items[^1];
var x = items[items.Count-1];
int[42] items;
int x = items[$-1];
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()
x = last items
foo :: [a] -> Maybe a
foo [] = Nothing
foo xs = Just $ last xs

x = foo 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]
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[items.len()-1];
let x = items.last().unwrap();
val x = items.last
x := items last.
Dim x = items(items.Count - 1)