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Programming-Idioms

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  • Perl

Idiom #22 Convert string to integer

Extract the integer value i from its string representation s (in radix 10)

Turning the string \"123\" into the integer value 123 
my $i = $s + 0;

Perl automatically converts numbers to strings, and strings to numbers, whenever required. This means the addition of 0 is strictly superfluous, it only indicates programmer intent.

+ always is addition, not concatenation.
I : Integer := Integer'Value (s);
New implementation...
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my $i = $s + 0;

I : Integer := Integer'Value (s);
i = (int)strtol(s, (char **)NULL, 10);
int i = atoi(s);
(def i (Integer/parseInt s))
(def i (Integer. s))
std::string s("123");
int i;
std::from_chars(s.data(), s.data() + s.size(), i, 10);
int i = std::atoi(s);
std::stringstream ss(str);
int i;
ss >> i;
int i = std::stoi(s);
long i = Convert.ToInt64(s);
int i = int.Parse(s);
auto i = s.to!int;
var i = int.parse(s);
i = String.to_integer(s)
I = list_to_integer(S).
read (unit=s,fmt=*) i
i, err  := strconv.Atoi(s) 
i, err := strconv.ParseInt(s, 10, 0)
Integer i = s.toInteger()
let i = read s :: Integer
const i = +s
let i = parseInt(s, 10)
const i = Number(s);
int i = Integer.parseInt(s);
Integer i = Integer.valueOf(s, 10);
int i = s.chars()
     .map(x -> x - '0')
     .reduce((a, b) -> (a * 10) + b)
     .getAsInt();
int i = new Integer(s).intValue();
val i = s.toInt()
val i = s.toIntOrNull()
(setf i (parse-integer s))
i = tonumber(s)
int i=s.intValue
$i = intval($s, 10);
i := StrToInt(S);
i = int(s)
i = s.to_i
let i = match s.parse::<i32>() {
  Ok(i) => i,
  Err(_e) => -1,
};
let i: i32 = s.parse().unwrap_or(0);
let i = s.parse::<i32>().unwrap();
s.toInt
(define i (string->number s))
i := s asInteger.
Dim myint As Integer = CInt(("123"))