Idiom #22 Convert string to integer

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

let i: i32 = s.parse().unwrap_or(0);

let i = match s.parse::<i32>() {
  Ok(i) => i,
  Err(_e) => -1,
};

let i = s.parse::<i32>().unwrap();

I : Integer := Integer'Value (s);

#include <stdlib.h>

i = (int)strtol(s, (char **)NULL, 10);

#include <stdlib.h>

int i = atoi(s);

(def i (Integer. s))

(def i (Integer/parseInt s))

#include <string>
#include <sstream>

std::stringstream ss(str);
int i;
ss >> i;

std::string s("123");
int i;
std::from_chars(s.data(), s.data() + s.size(), i, 10);

#include <cstdlib>

int i = std::atoi(s);

#include <string>

int i = std::stoi(s);

long i = Convert.ToInt64(s);

int i = int.Parse(s);

import std.conv;

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

import "strconv"

i, err  := strconv.Atoi(s)

import "strconv"

i, err := strconv.ParseInt(s, 10, 0)

Integer i = s.toInteger()

let i = read s :: Integer

let i = parseInt(s, 10)

const i = Number(s);

const i = +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)

@import Foundation;

int i=s.intValue

$i = intval($s, 10);

uses SysUtils;

i := StrToInt(S);

my $i = $s + 0;

i = int(s)

i = s.to_i

s.toInt

(define i (string->number s))

i := s asInteger.

Dim myint As Integer = CInt(("123"))

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Idiom created by programming-idioms.org

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