How do I convert a string to an integer in JavaScript?
The simplest way would be to use the native Number function:
var x = Number("1000")
If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.
parseInt()
var x = parseInt("1000", 10); // You want to use radix 10
// So you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])
Unary plus
If your string is already in the form of an integer:
var x = +"1000";
floor()
If your string is or might be a float and you want an integer:
var x = Math.floor("1000.01"); // floor() automatically converts string to number
Or, if you're going to be using Math.floor several times:
var floor = Math.floor;
var x = floor("1000.01");
parseFloat()
If you're the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.
var floor = Math.floor;
var x = floor(parseFloat("1000.01"));
round()
Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:
var round = Math.round;
var x = round("1000"); // Equivalent to round("1000", 0)
Try parseInt function:
var number = parseInt("10");
But there is a problem. If you try to convert "010" using parseInt function, it detects as octal number, and will return number 8. So, you need to specify a radix (from 2 to 36). In this case base 10.
parseInt(string, radix)
Example:
var result = parseInt("010", 10) == 10; // Returns true
var result = parseInt("010") == 10; // Returns false
Note that parseInt ignores bad data after parsing anything valid.
This guid will parse as 51:
var result = parseInt('51e3daf6-b521-446a-9f5b-a1bb4d8bac36', 10) == 51; // Returns true
There are two main ways to convert a string to a number in JavaScript. One way is to parse it and the other way is to change its type to a Number. All of the tricks in the other answers (e.g., unary plus) involve implicitly coercing the type of the string to a number. You can also do the same thing explicitly with the Number function.
Parsing
var parsed = parseInt("97", 10);
parseInt and parseFloat are the two functions used for parsing strings to numbers. Parsing will stop silently if it hits a character it doesn't recognise, which can be useful for parsing strings like "92px", but it's also somewhat dangerous, since it won't give you any kind of error on bad input, instead you'll get back NaN unless the string starts with a number. Whitespace at the beginning of the string is ignored. Here's an example of it doing something different to what you want, and giving no indication that anything went wrong:
var widgetsSold = parseInt("97,800", 10); // widgetsSold is now 97
It's good practice to always specify the radix as the second argument. In older browsers, if the string started with a 0, it would be interpreted as octal if the radix wasn't specified which took a lot of people by surprise. The behaviour for hexadecimal is triggered by having the string start with 0x if no radix is specified, e.g., 0xff. The standard actually changed with ECMAScript 5, so modern browsers no longer trigger octal when there's a leading 0 if no radix has been specified. parseInt understands radixes up to base 36, in which case both upper and lower case letters are treated as equivalent.
Changing the Type of a String to a Number
All of the other tricks mentioned above that don't use parseInt, involve implicitly coercing the string into a number. I prefer to do this explicitly,
var cast = Number("97");
This has different behavior to the parse methods (although it still ignores whitespace). It's more strict: if it doesn't understand the whole of the string than it returns NaN, so you can't use it for strings like 97px. Since you want a primitive number rather than a Number wrapper object, make sure you don't put new in front of the Number function.
Obviously, converting to a Number gives you a value that might be a float rather than an integer, so if you want an integer, you need to modify it. There are a few ways of doing this:
var rounded = Math.floor(Number("97.654")); // other options are Math.ceil, Math.round
var fixed = Number("97.654").toFixed(0); // rounded rather than truncated
var bitwised = Number("97.654")|0; // do not use for large numbers
Any bitwise operator (here I've done a bitwise or, but you could also do double negation as in an earlier answer or a bit shift) will convert the value to a 32 bit integer, and most of them will convert to a signed integer. Note that this will not do want you want for large integers. If the integer cannot be represented in 32 bits, it will wrap.
~~"3000000000.654" === -1294967296
// This is the same as
Number("3000000000.654")|0
"3000000000.654" >>> 0 === 3000000000 // unsigned right shift gives you an extra bit
"300000000000.654" >>> 0 === 3647256576 // but still fails with larger numbers
To work correctly with larger numbers, you should use the rounding methods
Math.floor("3000000000.654") === 3000000000
// This is the same as
Math.floor(Number("3000000000.654"))
Bear in mind that coercion understands exponential notation and Infinity, so 2e2 is 200 rather than NaN, while the parse methods don't.
Custom
It's unlikely that either of these methods do exactly what you want. For example, usually I would want an error thrown if parsing fails, and I don't need support for Infinity, exponentials or leading whitespace. Depending on your use case, sometimes it makes sense to write a custom conversion function.
Always check that the output of Number or one of the parse methods is the sort of number you expect. You will almost certainly want to use isNaN to make sure the number is not NaN (usually the only way you find out that the parse failed).
ParseInt() and + are different
parseInt("10.3456") // returns 10
+"10.3456" // returns 10.3456
Fastest
var x = "1000"*1;
Test
Here is little comparison of speed (macOS only)... :)
For Chrome, 'plus' and 'mul' are fastest (>700,000,00 op/sec), 'Math.floor' is slowest. For Firefox, 'plus' is slowest (!) 'mul' is fastest (>900,000,000 op/sec). In Safari 'parseInt' is fastest, 'number' is slowest (but results are quite similar, >13,000,000 <31,000,000). So Safari for cast string to int is more than 10x slower than other browsers. So the winner is 'mul' :)
You can run it on your browser by this link
https://jsperf.com/js-cast-str-to-number/1
I also tested var x = ~~"1000";. On Chrome and Safari, it is a little bit slower than var x = "1000"*1 (<1%), and on Firefox it is a little bit faster (<1%).
I use this way of converting string to number:
var str = "25"; // String
var number = str*1; // Number
So, when multiplying by 1, the value does not change, but JavaScript automatically returns a number.
But as it is shown below, this should be used if you are sure that the str is a number (or can be represented as a number), otherwise it will return NaN - not a number.
You can create simple function to use, e.g.,
function toNumber(str) {
return str*1;
}
Try parseInt.
var number = parseInt("10", 10); //number will have value of 10.
I love this trick:
~~"2.123"; //2
~~"5"; //5
The double bitwise negative drops off anything after the decimal point AND converts it to a number format. I've been told it's slightly faster than calling functions and whatnot, but I'm not entirely convinced.
Another method I just saw here (a question about the JavaScript >>> operator, which is a zero-fill right shift) which shows that shifting a number by 0 with this operator converts the number to a uint32 which is nice if you also want it unsigned. Again, this converts to an unsigned integer, which can lead to strange behaviors if you use a signed number.
"-2.123" >>> 0; // 4294967294
"2.123" >>> 0; // 2
"-5" >>> 0; // 4294967291
"5" >>> 0; // 5
In JavaScript, you can do the following:
ParseInt
parseInt("10.5") // Returns 10
Multiplying with 1
var s = "10";
s = s*1; // Returns 10
Using the unary operator (+)
var s = "10";
s = +s; // Returns 10
Using a bitwise operator
(Note: It starts to break after 2140000000. Example: ~~"2150000000" = -2144967296)
var s = "10.5";
s = ~~s; // Returns 10
Using Math.floor() or Math.ceil()
var s = "10";
s = Math.floor(s) || Math.ceil(s); // Returns 10
Please see the below example. It will help answer your question.
Example Result
parseInt("4") 4
parseInt("5aaa") 5
parseInt("4.33333") 4
parseInt("aaa"); NaN (means "Not a Number")
By using parseint function, it will only give op of integer present and not the string.
Beware if you use parseInt to convert a float in scientific notation!
For example:
parseInt("5.6e-14")
will result in
5
instead of
0
Also as a side note: MooTools has the function toInt() which is used on any native string (or float (or integer)).
"2".toInt() // 2
"2px".toInt() // 2
2.toInt() // 2
We can use +(stringOfNumber) instead of using parseInt(stringOfNumber).
Example: +("21") returns int of 21, like the parseInt("21").
We can use this unary "+" operator for parsing float too...
To convert a String into Integer, I recommend using parseFloat and not parseInt. Here's why:
Using parseFloat:
parseFloat('2.34cms') //Output: 2.34
parseFloat('12.5') //Output: 12.5
parseFloat('012.3') //Output: 12.3
Using parseInt:
parseInt('2.34cms') //Output: 2
parseInt('12.5') //Output: 12
parseInt('012.3') //Output: 12
So if you have noticed parseInt discards the values after the decimals, whereas parseFloat lets you work with floating point numbers and hence more suitable if you want to retain the values after decimals. Use parseInt if and only if you are sure that you want the integer value.
There are many ways in JavaScript to convert a string to a number value... All are simple and handy. Choose the way which one works for you:
var num = Number("999.5"); //999.5
var num = parseInt("999.5", 10); //999
var num = parseFloat("999.5"); //999.5
var num = +"999.5"; //999.5
Also, any Math operation converts them to number, for example...
var num = "999.5" / 1; //999.5
var num = "999.5" * 1; //999.5
var num = "999.5" - 1 + 1; //999.5
var num = "999.5" - 0; //999.5
var num = Math.floor("999.5"); //999
var num = ~~"999.5"; //999
My prefer way is using + sign, which is the elegant way to convert a string to number in JavaScript.
Try str - 0 to convert string to number.
> str = '0'
> str - 0
0
> str = '123'
> str - 0
123
> str = '-12'
> str - 0
-12
> str = 'asdf'
> str - 0
NaN
> str = '12.34'
> str - 0
12.34
Here are two links to compare the performance of several ways to convert string to int
https://jsperf.com/number-vs-parseint-vs-plus
http://phrogz.net/js/string_to_number.html
Here is the easiest solution
let myNumber = "123" | 0;
More easy solution
let myNumber = +"123";
In my opinion, no answer covers all edge cases as parsing a float should result in an error.
function parseInteger(value) {
if(value === '') return NaN;
const number = Number(value);
return Number.isInteger(number) ? number : NaN;
}
parseInteger("4") // 4
parseInteger("5aaa") // NaN
parseInteger("4.33333") // NaN
parseInteger("aaa"); // NaN
The easiest way would be to use + like this
const strTen = "10"
const numTen = +strTen // string to number conversion
console.log(typeof strTen) // string
console.log(typeof numTen) // number
I actually needed to "save" a string as an integer, for a binding between C and JavaScript, so I convert the string into an integer value:
/*
Examples:
int2str( str2int("test") ) == "test" // true
int2str( str2int("t€st") ) // "t¬st", because "€".charCodeAt(0) is 8364, will be AND'ed with 0xff
Limitations:
maximum 4 characters, so it fits into an integer
*/
function str2int(the_str) {
var ret = 0;
var len = the_str.length;
if (len >= 1) ret += (the_str.charCodeAt(0) & 0xff) << 0;
if (len >= 2) ret += (the_str.charCodeAt(1) & 0xff) << 8;
if (len >= 3) ret += (the_str.charCodeAt(2) & 0xff) << 16;
if (len >= 4) ret += (the_str.charCodeAt(3) & 0xff) << 24;
return ret;
}
function int2str(the_int) {
var tmp = [
(the_int & 0x000000ff) >> 0,
(the_int & 0x0000ff00) >> 8,
(the_int & 0x00ff0000) >> 16,
(the_int & 0xff000000) >> 24
];
var ret = "";
for (var i=0; i<4; i++) {
if (tmp[i] == 0)
break;
ret += String.fromCharCode(tmp[i]);
}
return ret;
}
String to Number in JavaScript:
Unary + (most recommended)
+numStr is easy to use and has better performance compared with others
Supports both integers and decimals
console.log(+'123.45') // => 123.45
Some other options:
Parsing Strings:
parseInt(numStr) for integers
parseFloat(numStr) for both integers and decimals
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
JavaScript Functions
Math functions like round(numStr), floor(numStr), ceil(numStr) for integers
Number(numStr) for both integers and decimals
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
Unary Operators
All basic unary operators, +numStr, numStr-0, 1*numStr, numStr*1, and numStr/1
All support both integers and decimals
Be cautious about numStr+0. It returns a string.
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // =>3.3
console.log('123.123'+0, typeof ('123.123' + 0)) // => 123.1230 string
Bitwise Operators
Two tilde ~~numStr or left shift 0, numStr<<0
Supports only integers, but not decimals
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
// Parsing
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
// Function
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
// Unary
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // => 3.3
console.log('123.123'+0, typeof ('123.123'+0)) // => 123.1230 string
// Bitwise
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
function parseIntSmarter(str) {
// ParseInt is bad because it returns 22 for "22thisendsintext"
// Number() is returns NaN if it ends in non-numbers, but it returns 0 for empty or whitespace strings.
return isNaN(Number(str)) ? NaN : parseInt(str, 10);
}
You can use plus.
For example:
var personAge = '24';
var personAge1 = (+personAge)
then you can see the new variable's type bytypeof personAge1 ; which is number.
Summing the multiplication of digits with their respective power of ten:
i.e: 123 = 100+20+3 = 1100 + 2+10 + 31 = 1*(10^2) + 2*(10^1) + 3*(10^0)
function atoi(array) {
// Use exp as (length - i), other option would be
// to reverse the array.
// Multiply a[i] * 10^(exp) and sum
let sum = 0;
for (let i = 0; i < array.length; i++) {
let exp = array.length - (i+1);
let value = array[i] * Math.pow(10, exp);
sum += value;
}
return sum;
}
The safest way to ensure you get a valid integer:
let integer = (parseInt(value, 10) || 0);
Examples:
// Example 1 - Invalid value:
let value = null;
let integer = (parseInt(value, 10) || 0);
// => integer = 0
// Example 2 - Valid value:
let value = "1230.42";
let integer = (parseInt(value, 10) || 0);
// => integer = 1230
// Example 3 - Invalid value:
let value = () => { return 412 };
let integer = (parseInt(value, 10) || 0);
// => integer = 0
Another option is to double XOR the value with itself:
var i = 12.34;
console.log('i = ' + i);
console.log('i ⊕ i ⊕ i = ' + (i ^ i ^ i));
This will output:
i = 12.34
i ⊕ i ⊕ i = 12
I only added one plus(+) before string and that was solution!
+"052254" // 52254
Number()
Number(" 200.12 ") // Returns 200.12
Number("200.12") // Returns 200.12
Number("200") // Returns 200
parseInt()
parseInt(" 200.12 ") // Return 200
parseInt("200.12") // Return 200
parseInt("200") // Return 200
parseInt("Text information") // Returns NaN
parseFloat()
It will return the first number
parseFloat("200 400") // Returns 200
parseFloat("200") // Returns 200
parseFloat("Text information") // Returns NaN
parseFloat("200.10") // Return 200.10
Math.floor()
Round a number to the nearest integer
Math.floor(" 200.12 ") // Return 200
Math.floor("200.12") // Return 200
Math.floor("200") // Return 200
function doSth(){
var a = document.getElementById('input').value;
document.getElementById('number').innerHTML = toNumber(a) + 1;
}
function toNumber(str){
return +str;
}
<input id="input" type="text">
<input onclick="doSth()" type="submit">
<span id="number"></span>
This (probably) isn't the best solution for parsing an integer, but if you need to "extract" one, for example:
"1a2b3c" === 123
"198some text2hello world!30" === 198230
// ...
this would work (only for integers):
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (!isNaN(str[i - 1])) {
result += parseInt(str[i - 1]) * factor
factor *= 10
}
}
return result
}
console.log(extractInteger(str))
Of course, this would also work for parsing an integer, but would be slower than other methods.
You could also parse integers with this method and return NaN if the string isn't a number, but I don't see why you'd want to since this relies on parseInt internally and parseInt is probably faster.
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (isNaN(str[i - 1])) return NaN
result += parseInt(str[i - 1]) * factor
factor *= 10
}
return result
}
console.log(extractInteger(str))
I need to find the value (or position) of the most significant bit (MSB) of an integer in Swift.
Eg:
Input number: 9
Input as binary: 1001
MS value as binary: 1000 -> (which is 8 in decimal)
MS position as decimal: 3 (because 1<<3 == 1000)
Many processors (Intel, AMD, ARM) have instructions for this. In c, these are exposed. Are these instructions similarly available in Swift through a library function, or would I need to implement some bit twiddling?
The value is more useful in my case.
If a position is returned, then the value can be easily derived by a single shift.
Conversely, computing position from value is not so easy unless a fast Hamming Weight / pop count function is available.
You can use the flsl() function ("find last set bit, long"):
let x = 9
let p = flsl(x)
print(p) // 4
The result is 4 because flsl() and the related functions number the bits starting at 1, the least significant bit.
On Intel platforms you can use the _bit_scan_reverse intrinsic,
in my test in a macOS application this translated to a BSR
instruction.
import _Builtin_intrinsics.intel
let x: Int32 = 9
let p = _bit_scan_reverse(x)
print(p) // 3
You can use the the properties leadingZeroBitCount and trailingZeroBitCount to find the Most Significant Bit and Least Significant Bit.
For example,
let i: Int = 95
let lsb = i.trailingZeroBitCount
let msb = Int.bitWidth - 1 - i.leadingZeroBitCount
print("i: \(i) = \(String(i, radix: 2))") // i: 95 = 1011111
print("lsb: \(lsb) = \(String(1 << lsb, radix: 2))") // lsb: 0 = 1
print("msb: \(msb) = \(String(1 << msb, radix: 2))") // msb: 6 = 1000000
If you look at the disassembly(ARM Mac) in LLDB for the Least Significant Bit code, it uses a single instruction, clz, to count the zeroed bits. (ARM Reference)
** 15 let lsb = i.trailingZeroBitCount
0x100ed947c <+188>: rbit x9, x8
0x100ed9480 <+192>: clz x9, x9
0x100ed9484 <+196>: mov x10, x9
0x100ed9488 <+200>: str x10, [sp, #0x2d8]
I'm looking to create a function that returns a solve for x math equation that can be preformed in ones head (Clearly thats a bit subjective but I'm not sure how else to phrase it).
Example problem: (x - 15)/10 = 6
Note: Only 1 x in the equation
I want to use the operations +, -, *, /, sqrt (Only applied to X -> sqrt(x))
I know that let mathExpression = NSExpression(format: question) converts strings into math equations but when solving for x I'm not sure how to go about doing this.
I previously asked Generating random doable math problems swift for non solving for x problems but I'm not sure how to convert that answer into solving for x
Edit: Goal is to generate an equation and have the user solve for the variable.
Since all you want is a string representing an equation and a value for x, you don't need to do any solving. Just start with x and transform it until you have a nice equation. Here's a sample: (copy and paste it into a Playground to try it out)
import UIKit
enum Operation: String {
case addition = "+"
case subtraction = "-"
case multiplication = "*"
case division = "/"
static func all() -> [Operation] {
return [.addition, .subtraction, .multiplication, .division]
}
static func random() -> Operation {
let all = Operation.all()
let selection = Int(arc4random_uniform(UInt32(all.count)))
return all[selection]
}
}
func addNewTerm(formula: String, result: Int) -> (formula: String, result: Int) {
// choose a random number and operation
let operation = Operation.random()
let number = chooseRandomNumberFor(operation: operation, on: result)
// apply to the left side
let newFormula = applyTermTo(formula: formula, number: number, operation: operation)
// apply to the right side
let newResult = applyTermTo(result: result, number: number, operation: operation)
return (newFormula, newResult)
}
func applyTermTo(formula: String, number:Int, operation:Operation) -> String {
return "\(formula) \(operation.rawValue) \(number)"
}
func applyTermTo(result: Int, number:Int, operation:Operation) -> Int {
switch(operation) {
case .addition: return result + number
case .subtraction: return result - number
case .multiplication: return result * number
case .division: return result / number
}
}
func chooseRandomNumberFor(operation: Operation, on number: Int) -> Int {
switch(operation) {
case .addition, .subtraction, .multiplication:
return Int(arc4random_uniform(10) + 1)
case .division:
// add code here to find integer factors
return 1
}
}
func generateFormula(_ numTerms:Int = 1) -> (String, Int) {
let x = Int(arc4random_uniform(10))
var leftSide = "x"
var result = x
for i in 1...numTerms {
(leftSide, result) = addNewTerm(formula: leftSide, result: result)
if i < numTerms {
leftSide = "(" + leftSide + ")"
}
}
let formula = "\(leftSide) = \(result)"
return (formula, x)
}
func printFormula(_ numTerms:Int = 1) {
let (formula, x) = generateFormula(numTerms)
print(formula, " x = ", x)
}
for i in 1...30 {
printFormula(Int(arc4random_uniform(3)) + 1)
}
There are some things missing. The sqrt() function will have to be implemented separately. And for division to be useful, you'll have to add in a system to find factors (since you presumably want the results to be integers). Depending on what sort of output you want, there's a lot more work to do, but this should get you started.
Here's sample output:
(x + 10) - 5 = 11 x = 6
((x + 6) + 6) - 1 = 20 x = 9
x - 2 = 5 x = 7
((x + 3) * 5) - 6 = 39 x = 6
(x / 1) + 6 = 11 x = 5
(x * 6) * 3 = 54 x = 3
x * 9 = 54 x = 6
((x / 1) - 6) + 8 = 11 x = 9
Okay, let’s assume from you saying “Note: Only 1 x in the equation” that what you want is a linear equation of the form y = 0 = β1*x + β0, where β0 and β1 are the slope and intercept coefficients, respectively.
The inverse of (or solution to) any linear equation is given by x = -β0/β1. So what you really need to do is generate random integers β0 and β1 to create your equation. But since it should be “solvable” in someone’s head, you probably want β0 to be divisible by β1, and furthermore, for β1 and β0/β1 to be less than or equal to 12, since this is the upper limit of the commonly known multiplication tables. In this case, just generate a random integer β1 ≤ 12, and β0 equal to β1 times some integer n, 0 ≤ n ≤ 12.
If you want to allow simple fractional solutions like 2/3, just multiply the denominator and the numerator into β0 and β1, respectively, taking care to prevent the numerator or denominator from getting too large (12 is again a good limit).
Since you probably want to make y non-zero, just generate a third random integer y between -12 and 12, and change your output equation to y = β1*x + β0 + y.
Since you mentioned √ could occur over the x variable only, that is pretty easy to add; the solution (to 0 = β1*sqrt(x) + β0) is just x = (β0/β1)**2.
Here is some very simple (and very problematic) code for generating random integers to get you started:
import func Glibc.srand
import func Glibc.rand
import func Glibc.time
srand(UInt32(time(nil)))
print(rand() % 12)
There are a great many answers on this website that deal with better ways to generate random integers.