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Programming-Idioms

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  • Lisp

Idiom #38 Extract a substring

Find substring t consisting in characters i (included) to j (excluded) of string s.
Character indices start at 0 unless specified otherwise.
Make sure that multibyte characters are properly handled.

(setf u (subseq s i j))

t is not a good choice for a variable name in Lisp, so using u instead for the result string
T : String := S (I .. J - 1);

In Ada, strings use 1-based indexing

In Ada, J is included by default when slicing
New implementation...
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programming-idioms.org 
(setf u (subseq s i j))

T : String := S (I .. J - 1);
char *t=malloc((j-i+1)*sizeof(char));
strncpy(t,s+i,j-i);
(def t (subs s i j))
auto t = s.substr(i, j-i);
var t = s.Substring(i, j - i);
auto t = s[i .. j];
var t = s.substring(i, j);
t = String.slice(s, i..j-1)
T = string:sub_string(I, J-1).
character(len=:), allocatable :: t

  t = s(i:j-1)

t := string([]rune(s)[i:j])
def t = s[i..<j]
t = drop i (take j s)

let t = s.substring(i, j);
let t = s.slice(i, j);
String t = s.substring(i,j);

val t = s.substring(i, j)
local t = s:sub(i, j - 1)
t=[s substringWithRange:NSMakeRange(i,j-i)]
$t = mb_substr($s, $i, $j-$i, 'UTF-8');
t := copy(s,i+1,j-i);
t := s.Substring(i,j-i);
my $chunk = substr("now is the time", $i, $j);
t = s[slice(i, j)]
t = s[i:j]
t = s[i..j-1] 
let t = s.substring(i, j);
let t = s.graphemes(true).skip(i).take(j - i).collect::<String>();
let mut iter = s.grapheme_indices(true);
let i_idx = iter.nth(i).map(|x|x.0).unwrap_or(0);
let j_idx = iter.nth(j-i).map(|x|x.0).unwrap_or(0);
let t = s[i_idx..j_idx];
val t = s.substring(i, j)
(define t (substring s i j))
t := s copyFrom: i to: j - 1.
Dim t As String = s.Substring(i,j)