I am pretty new at this stuff, and I am solving a series of questions, but I've got lost in this one.
I have to verify if a number can be divided by other, and the answer must be true or false.
I got his
function solucao(numero, x) {
solucao(numero % x);
if (solucao === 0) {
resultado = true;
}else{
resultado = false;
}
}
But I'm getting runtime error and can't see whats is missing.
So you want to check if a number numero is divisible by x. The modulo operator can help. Try this:
function solucao(numero, x){
if (numero % x == 0){
return true
}
else {
return false
}
}
function solucao(numero, x) {
let resultado;
if (numero % x === 0) {
resultado = true;
}else{
resultado = false;
}
return resultado;
}
I think you get confused at some point. You are calling the function, inside of itself. You should do like this, and also, declare the result variable.
I am sure this will help:
function checkIfDivided(){
// in this section the variables come from an html document
var number=parseInt(document.getElementById("number").value);
var divisor=parseInt(document.getElementById("divisor").value);
if(number%divisor==0)
return true;
else return false;
}
or
function checkIfDivided(number,divisor){
//in the function the variable are given as parameters
if(number%divisor==0)
return true;
else return false;
}
Looks like two things to me:
You haven't declared your 'resultado' variable ( this can be as simple as just typing 'let resultado;' without the single quotes
You haven't returned your 'resultado' variable after the if/else statement
Right now, your function is using an undeclared variable and not returning anything, so that is why you are getting an error. Fix the two above steps and you should be good! :)
You clearly understand that the modulus operator is the way to go. Using it we discover that 12 is divisible by 3 because 12 % 3 return zero. Zero is considered a "falsy" value while any other number is considered "truthy".
Given this then if 12 % 3 returns a "falsey" value (zero) we can't use the result directly. But what if we can "flip" false to true? We can, using the not operator (!).
Using the ! operator on the result of a math problem requires the use of parentheses around the math problem itself.
So the problem in code becomes (12 % 3) and to 'flip' it with the ! operator it becomes
!(12 % 3).
This is proven below with:
console.log(!(12 % 3)) --> logs true
console.log(!(12 % 5)) --> logs false
The function implementation of that is simple and also proven:
console.log(isDivisible(12,3)); --> logs true
console.log(isDivisible(12,5)); --> logs false
console.log(!(12 % 3))
console.log(!(12 % 5))
function isDivisible(number, x){
return !(number % x);
}
console.log(isDivisible(12,3));
console.log(isDivisible(12,5));
There is one other way to do so and i think its much cleaner.
console.log(Number.isInteger(10/2)) //true
console.log(Number.isInteger(4/2)) // false
//a must be greater than b
function check(a,b) {
console.log(Number.isInteger(a/b))
return Number.isInteger(a/b)
}
check(10,5)//true
check(8,3)//false
I make a project that reads value from a remote JSON. In this project the remote JSON can give me with one variable 16 different types of alarm. The implementation is by a bynary 16bit value expressed in int. So if there is the third alarm it should return me 8 (bynary 1000) and if the second,eigth and tenth alarm is up it return me 1284 (binary 10100000100). Well when there is no alarm it returns me 0.
So, I create a function in JS (accordly to here) that passing the value returned (8/1284/0 in the example) returns me a simple true or false if there is an alarm. The function is pretty simple:
function IsOnAlarm(passedVal) {
if (passedVal & 0) {
return false;
} else {
return true;
}
}
but it does not function :-(
I create a JSFiddle that shows the problem. How can I solve it? Thanks in advance.
Well as far as I understand you just need to check whether the value is 0. There's no need for any bitwise operation.
function IsOnAlarm(passedVal) {
return passedVal != 0;
}
Side note: passedVal & 0 is always 0 no matter what passedVal is (that's why it always goes to else).
You can do it like
function IsOnAlarm(passedVal) {
if (passedVal != 0) {
return false;
} else {
return true;
}
}
Check this http://jsfiddle.net/KQH43/3/
The AND operator always return false if one of the operands is zero, see below:
0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
1 & 1 = 1
The easiest way to rewrite your function is as follows:
function IsOnAlarm(passedVal) {
return !!passedVal;
}
Explanation: the double negation operation (!!) allows you to cast the input to boolean. Any non-zero value will evaluate to true, zero evaluates to false.
Check this jsfiddle, it is using bitwise AND comparison.
I've modified your function to make it easier to understand:
function AreAlarmsOn(alarmsToCheck, alarmState) {
return and(alarmsToCheck, alarmState)!=0;
}
It takes alarmsToCheck and checks its bits against alarmState bits, alarmsToCheck can have bits for several alarms if you need to check multiple alarms.
Is this what you're looking for?
var getFlags = function(val) {
var results = [];
for (var i = 0; i < 16; i++) {
results.push(!!((val >> i) & 1));
}
return results;
};
var flags = getFlags(1284);
flags[2] //=> true
flags[8] //=> true
flags[10] //=> true
// all other values ==> false / undefined
How do I check if a variable is an integer in JavaScript, and throw an alert if it isn't? I tried this, but it doesn't work:
<html>
<head>
<script type="text/javascript">
var data = 22;
alert(NaN(data));
</script>
</head>
</html>
That depends, do you also want to cast strings as potential integers as well?
This will do:
function isInt(value) {
return !isNaN(value) &&
parseInt(Number(value)) == value &&
!isNaN(parseInt(value, 10));
}
With Bitwise operations
Simple parse and check
function isInt(value) {
var x = parseFloat(value);
return !isNaN(value) && (x | 0) === x;
}
Short-circuiting, and saving a parse operation:
function isInt(value) {
if (isNaN(value)) {
return false;
}
var x = parseFloat(value);
return (x | 0) === x;
}
Or perhaps both in one shot:
function isInt(value) {
return !isNaN(value) && (function(x) { return (x | 0) === x; })(parseFloat(value))
}
Tests:
isInt(42) // true
isInt("42") // true
isInt(4e2) // true
isInt("4e2") // true
isInt(" 1 ") // true
isInt("") // false
isInt(" ") // false
isInt(42.1) // false
isInt("1a") // false
isInt("4e2a") // false
isInt(null) // false
isInt(undefined) // false
isInt(NaN) // false
Here's the fiddle: http://jsfiddle.net/opfyrqwp/28/
Performance
Testing reveals that the short-circuiting solution has the best performance (ops/sec).
// Short-circuiting, and saving a parse operation
function isInt(value) {
var x;
if (isNaN(value)) {
return false;
}
x = parseFloat(value);
return (x | 0) === x;
}
Here is a benchmark:
http://jsben.ch/#/htLVw
If you fancy a shorter, obtuse form of short circuiting:
function isInt(value) {
var x;
return isNaN(value) ? !1 : (x = parseFloat(value), (0 | x) === x);
}
Of course, I'd suggest letting the minifier take care of that.
Use the === operator (strict equality) as below,
if (data === parseInt(data, 10))
alert("data is integer")
else
alert("data is not an integer")
Number.isInteger() seems to be the way to go.
MDN has also provided the following polyfill for browsers not supporting Number.isInteger(), mainly all versions of IE.
Link to MDN page
Number.isInteger = Number.isInteger || function(value) {
return typeof value === "number" &&
isFinite(value) &&
Math.floor(value) === value;
};
Assuming you don't know anything about the variable in question, you should take this approach:
if(typeof data === 'number') {
var remainder = (data % 1);
if(remainder === 0) {
// yes, it is an integer
}
else if(isNaN(remainder)) {
// no, data is either: NaN, Infinity, or -Infinity
}
else {
// no, it is a float (still a number though)
}
}
else {
// no way, it is not even a number
}
To put it simply:
if(typeof data==='number' && (data%1)===0) {
// data is an integer
}
You could check if the number has a remainder:
var data = 22;
if(data % 1 === 0){
// yes it's an integer.
}
Mind you, if your input could also be text and you want to check first it is not, then you can check the type first:
var data = 22;
if(typeof data === 'number'){
// yes it is numeric
if(data % 1 === 0){
// yes it's an integer.
}
}
You can use a simple regular expression:
function isInt(value) {
var er = /^-?[0-9]+$/;
return er.test(value);
}
In ES6 2 new methods are added for Number Object.
In it Number.isInteger() method returns true if the argument is an integer, otherwise returns false.
Important Note: The method will also return true for floating point numbers that can be represented as integer. Eg: 5.0 (as it is exactly equal to 5 )
Example usage :
Number.isInteger(0); // true
Number.isInteger(1); // true
Number.isInteger(-100000); // true
Number.isInteger(99999999999999999999999); // true
Number.isInteger(0.1); // false
Number.isInteger(Math.PI); // false
Number.isInteger(NaN); // false
Number.isInteger(Infinity); // false
Number.isInteger(-Infinity); // false
Number.isInteger('10'); // false
Number.isInteger(true); // false
Number.isInteger(false); // false
Number.isInteger([1]); // false
Number.isInteger(5.0); // true
Number.isInteger(5.000000000000001); // false
Number.isInteger(5.0000000000000001); // true
First off, NaN is a "number" (yes I know it's weird, just roll with it), and not a "function".
You need to check both if the type of the variable is a number, and to check for integer I would use modulus.
alert(typeof data === 'number' && data%1 == 0);
Be careful while using
num % 1
empty string ('') or boolean (true or false) will return as integer. You might not want to do that
false % 1 // true
'' % 1 //true
Number.isInteger(data)
Number.isInteger(22); //true
Number.isInteger(22.2); //false
Number.isInteger('22'); //false
build in function in the browser. Dosnt support older browsers
Alternatives:
Math.round(num)=== num
However, Math.round() also will fail for empty string and boolean
To check if integer like poster wants:
if (+data===parseInt(data)) {return true} else {return false}
notice + in front of data (converts string to number), and === for exact.
Here are examples:
data=10
+data===parseInt(data)
true
data="10"
+data===parseInt(data)
true
data="10.2"
+data===parseInt(data)
false
The simplest and cleanest pre-ECMAScript-6 solution (which is also sufficiently robust to return false even if a non-numeric value such as a string or null is passed to the function) would be the following:
function isInteger(x) { return (x^0) === x; }
The following solution would also work, although not as elegant as the one above:
function isInteger(x) { return Math.round(x) === x; }
Note that Math.ceil() or Math.floor() could be used equally well (instead of Math.round()) in the above implementation.
Or alternatively:
function isInteger(x) { return (typeof x === 'number') && (x % 1 === 0); }
One fairly common incorrect solution is the following:
function isInteger(x) { return parseInt(x, 10) === x; }
While this parseInt-based approach will work well for many values of x, once x becomes quite large, it will fail to work properly. The problem is that parseInt() coerces its first parameter to a string before parsing digits. Therefore, once the number becomes sufficiently large, its string representation will be presented in exponential form (e.g., 1e+21). Accordingly, parseInt() will then try to parse 1e+21, but will stop parsing when it reaches the e character and will therefore return a value of 1. Observe:
> String(1000000000000000000000)
'1e+21'
> parseInt(1000000000000000000000, 10)
1
> parseInt(1000000000000000000000, 10) === 1000000000000000000000
false
Why hasnt anyone mentioned Number.isInteger() ?
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/isInteger
Works perfectly for me and solves the issue with the NaN beginning a number.
if(Number.isInteger(Number(data))){
//-----
}
Check if the variable is equal to that same variable rounded to an integer, like this:
if(Math.round(data) != data) {
alert("Variable is not an integer!");
}
ECMA-262 6.0 (ES6) standard include Number.isInteger function.
In order to add support for old browser I highly recommend using strong and community supported solution from:
https://github.com/paulmillr/es6-shim
which is pure ES6 JS polyfills library.
Note that this lib require es5-shim, just follow README.md.
The Accepted answer not worked for me as i needed to check for int/float and alphabet. so try this at will work for both int/float and alphabet check
function is_int(value){
if( (parseInt(value) % 1 === 0 )){
return true;
}else{
return false;
}
}
usage
is_int(44); // true
is_int("44"); // true
is_int(44.55); // true
is_int("44.55"); // true
is_int("aaa"); // false
You could use this function:
function isInteger(value) {
return (value == parseInt(value));
}
It will return true even if the value is a string containing an integer value.
So, the results will be:
alert(isInteger(1)); // true
alert(isInteger(1.2)); // false
alert(isInteger("1")); // true
alert(isInteger("1.2")); // false
alert(isInteger("abc")); // false
Use the | operator:
(5.3 | 0) === 5.3 // => false
(5.0 | 0) === 5.0 // => true
So, a test function might look like this:
var isInteger = function (value) {
if (typeof value !== 'number') {
return false;
}
if ((value | 0) !== value) {
return false;
}
return true;
};
Number.isInteger() is the best way if your browser support it, if not, I think there are so many ways to go:
function isInt1(value){
return (value^0) === value
}
or:
function isInt2(value){
return (typeof value === 'number') && (value % 1 === 0);
}
or:
function isInt3(value){
return parseInt(value, 10) === value;
}
or:
function isInt4(value){
return Math.round(value) === value;
}
now we can test the results:
var value = 1
isInt1(value) // return true
isInt2(value) // return true
isInt3(value) // return true
isInt4(value) // return true
var value = 1.1
isInt1(value) // return false
isInt2(value) // return false
isInt3(value) // return false
isInt4(value) // return false
var value = 1000000000000000000
isInt1(value) // return false
isInt2(value) // return true
isInt3(value) // return false
isInt4(value) // return true
var value = undefined
isInt1(value) // return false
isInt2(value) // return false
isInt3(value) // return false
isInt4(value) // return false
var value = '1' //number as string
isInt1(value) // return false
isInt2(value) // return false
isInt3(value) // return false
isInt4(value) // return false
So, all of these methods are works, but when the number is very big, parseInt and ^ operator would not works well.
You could tryNumber.isInteger(Number(value)) if value might be an integer in string form e.g var value = "23" and you want this to evaluate to true. Avoid trying Number.isInteger(parseInt(value)) because this won't always return the correct value. e.g if var value = "23abc" and you use the parseInt implementation, it would still return true.
But if you want strictly integer values then probably Number.isInteger(value) should do the trick.
var x = 1.5;
if(!isNaN(x)){
console.log('Number');
if(x % 1 == 0){
console.log('Integer');
}
}else {
console.log('not a number');
}
What about large integers (bigint)?
Most of these answers fail on large integers (253 and larger): Bitwise tests(e.g. (x | 0) === x), testing typeof x === 'number', regular int functions (e.g. parseInt), regular arithmetics fail on large integers. This can be resolved by using BigInt.
I've compiled several answers into one snippet to show the results. Most outright fail with large integers, while others work, except when passed the type BigInt (e.g. 1n). I've not included duplicate answers and have also left out any answers that allow decimals or don't attempt to test type)
// these all fail
n = 1000000000000000000000000000000
b = 1n
// These all fail on large integers
//https://stackoverflow.com/a/14636652/3600709
console.log('fail',1,n === parseInt(n, 10))
//https://stackoverflow.com/a/14794066/3600709
console.log('fail',2,!isNaN(n) && parseInt(Number(n)) == n && !isNaN(parseInt(n, 10)))
console.log('fail',2,!isNaN(n) && (parseFloat(n) | 0) === parseFloat(n))
console.log('fail',2,!isNaN(n) && (function(x) { return (x | 0) === x; })(parseFloat(n)))
//https://stackoverflow.com/a/21742529/3600709
console.log('fail',3,n == ~~n)
//https://stackoverflow.com/a/28211631/3600709
console.log('fail',4,!isNaN(n) && parseInt(n) == parseFloat(n))
//https://stackoverflow.com/a/41854178/3600709
console.log('fail',5,String(parseInt(n, 10)) === String(n))
// These ones work for integers, but not BigInt types (e.g. 1n)
//https://stackoverflow.com/a/14636725/3600709
console.log('partial',1,typeof n==='number' && (n%1)===0) // this one works
console.log('partial',1,typeof b==='number' && (b%1)===0) // this one fails
//https://stackoverflow.com/a/27424770/3600709
console.log('partial',2,Number.isInteger(n)) // this one works
console.log('partial',2,Number.isInteger(b)) // this one fails
//https://stackoverflow.com/a/14636638/3600709
console.log('partial',3,n % 1 === 0)
console.log('partial',3,b % 1 === 0) // gives uncaught type on BigInt
Checking type
If you actually want to test the incoming value's type to ensure it's an integer, use this instead:
function isInt(value) {
try {
BigInt(value)
return !['string','object','boolean'].includes(typeof value)
} catch(e) {
return false
}
}
function isInt(value) {
try {
BigInt(value)
return !['string','object','boolean'].includes(typeof value)
} catch(e) {
return false
}
}
console.log('--- should be false')
console.log(isInt(undefined))
console.log(isInt(''))
console.log(isInt(null))
console.log(isInt({}))
console.log(isInt([]))
console.log(isInt(1.1e-1))
console.log(isInt(1.1))
console.log(isInt(true))
console.log(isInt(NaN))
console.log(isInt('1'))
console.log(isInt(function(){}))
console.log(isInt(Infinity))
console.log('--- should be true')
console.log(isInt(10))
console.log(isInt(0x11))
console.log(isInt(0))
console.log(isInt(-10000))
console.log(isInt(100000000000000000000000000000000000000))
console.log(isInt(1n))
Without checking type
If you don't care if the incoming type is actually boolean, string, etc. converted into a number, then just use the following:
function isInt(value) {
try {
BigInt(value)
return true
} catch(e) {
return false
}
}
function isInt(value) {
try {
BigInt(value)
return true
} catch(e) {
return false
}
}
console.log('--- should be false')
console.log(isInt(undefined))
console.log(isInt(null))
console.log(isInt({}))
console.log(isInt(1.1e-1))
console.log(isInt(1.1))
console.log(isInt(NaN))
console.log(isInt(function(){}))
console.log(isInt(Infinity))
console.log('--- should be true')
console.log(isInt(10))
console.log(isInt(0x11))
console.log(isInt(0))
console.log(isInt(-10000))
console.log(isInt(100000000000000000000000000000000000000))
console.log(isInt(1n))
// gets converted to number
console.log(isInt(''))
console.log(isInt([]))
console.log(isInt(true))
console.log(isInt('1'))
My approach:
a >= 1e+21 → Only pass for very large numbers. This will cover all cases for sure, unlike other solutions which has been provided in this discussion.
a === (a|0) → if the given function's argument is exactly the same (===) as the bitwise-transformed value, it means that the argument is an integer.
a|0 → return 0 for any value of a that isn't a number, and if a is indeed a number, it will strip away anything after the decimal point, so 1.0001 will become 1
const isInteger = n => n >= 1e+21 ? true : n === (n|0);
// tests:
[
[1, true],
[1000000000000000000000, true],
[4e2, true],
[Infinity, true],
[1.0, true],
[1.0000000000001, false],
[0.1, false],
["0", false],
["1", false],
["1.1", false],
[NaN, false],
[[], false],
[{}, false],
[true, false],
[false, false],
[null, false],
[undefined, false],
].forEach(([test, expected]) =>
console.log(
isInteger(test) === expected,
typeof test,
test
)
)
Besides, Number.isInteger(). Maybe Number.isSafeInteger() is another option here by using the ES6-specified.
To polyfill Number.isSafeInteger(..) in pre-ES6 browsers:
Number.isSafeInteger = Number.isSafeInteger || function(num) {
return typeof num === "number" &&
isFinite(num) &&
Math.floor(num) === num &&
Math.abs( num ) <= Number.MAX_SAFE_INTEGER;
};
you can also try it this way
var data = 22;
if (Number.isInteger(data)) {
console.log("integer");
}else{
console.log("not an integer");
}
or
if (data === parseInt(data, 10)){
console.log("integer");
}else{
console.log("not an integer");
}
Just try this:
let number = 5;
if (Number.isInteger(number)) {
//do something
}
The 'accepted' answer is wrong (as some comments below point out).
this modification can make it work:
if (data.toString() === parseInt(data, 10).toString())
alert("data is a valid integer")
else
alert("data is not a valid integer")
You can use regexp for this:
function isInteger(n) {
return (typeof n == 'number' && /^-?\d+$/.test(n+''));
}
From http://www.toptal.com/javascript/interview-questions:
function isInteger(x) { return (x^0) === x; }
Found it to be the best way to do this.
function isInteger(argument) { return argument == ~~argument; }
Usage:
isInteger(1); // true<br>
isInteger(0.1); // false<br>
isInteger("1"); // true<br>
isInteger("0.1"); // false<br>
or:
function isInteger(argument) { return argument == argument + 0 && argument == ~~argument; }
Usage:
isInteger(1); // true<br>
isInteger(0.1); // false<br>
isInteger("1"); // false<br>
isInteger("0.1"); // false<br>