check for a remainder when dividing by 1:
function isInt(n) {
return n % 1 === 0;
}
If you don't know that the argument is a number you need two tests:
function isInt(n){
return Number(n) === n && n % 1 === 0;
}
function isFloat(n){
return Number(n) === n && n % 1 !== 0;
}
Update 2019 5 years after this answer was written, a solution was standardized in ECMA Script 2015. That solution is covered in this answer.
Answer from kennebec on Stack OverflowVideos
check for a remainder when dividing by 1:
function isInt(n) {
return n % 1 === 0;
}
If you don't know that the argument is a number you need two tests:
function isInt(n){
return Number(n) === n && n % 1 === 0;
}
function isFloat(n){
return Number(n) === n && n % 1 !== 0;
}
Update 2019 5 years after this answer was written, a solution was standardized in ECMA Script 2015. That solution is covered in this answer.
There is a method called Number.isInteger() which is currently implemented in everything but IE. MDN also provides a polyfill for other browsers:
Number.isInteger = Number.isInteger || function(value) {
return typeof value === 'number' &&
isFinite(value) &&
Math.floor(value) === value;
};
However, for most uses cases, you are better off using Number.isSafeInteger which also checks if the value is so high/low that any decimal places would have been lost anyway. MDN has a polyfil for this as well. (You also need the isInteger pollyfill above.)
if (!Number.MAX_SAFE_INTEGER) {
Number.MAX_SAFE_INTEGER = 9007199254740991; // Math.pow(2, 53) - 1;
}
Number.isSafeInteger = Number.isSafeInteger || function (value) {
return Number.isInteger(value) && Math.abs(value) <= Number.MAX_SAFE_INTEGER;
};
To convert a string to a Number you can use the Number constructor:
const n = Number('my string') // NaN ("not a number")
To check to see if the conversion was a valid number:
Number.isNaN(n)
Note: Internet Explorer does not support Number.isNaN and for that you can use the older isNaN. More here and here.
To check if a Number is an integer, there are two built-in methods:
Number.isInteger(), andNumber.isSafeInteger()
Further discussion can be found here.
let a = [4, 2.3, "SO2", 4, "O2", 3.4, 4.5, "CO2", 5.6, 3.4, 2];
a.forEach((e)=>{
const n = Number(e)
if (Number.isNaN(n)) {
console.log(e + " :letter")
return
}
if (Number.isInteger(n)) {
console.log(e + " :integer")
return
}
console.log(e + " :non-integer number")
})
Note
“Floating point” is the name of the technology used to represent the numbers in binary. This is why I changed the terminology from "float" to "non-integer number" in the code above.
Some numeric values in JavaScript contain non-numeric characters. eg.
Infinity
0x3d // hexadecimal
3n // BigInt
0b101011 // Binary
1e2 // times ten to the power of...
There are now two kinds of number in JavaScript: Number and BigInt.
Number is an implementation of the IEEE 754 32-bit floating point number specification.
This should Work
function isFloat(n){
return Number(n) === n && n % 1 !== 0;
}
var intvalue = Math.floor( floatvalue );
var intvalue = Math.ceil( floatvalue );
var intvalue = Math.round( floatvalue );
// `Math.trunc` was added in ECMAScript 6
var intvalue = Math.trunc( floatvalue );
Math object reference
Examples
Positive// value=x // x=5 5<x<5.5 5.5<=x<6
Math.floor(value) // 5 5 5
Math.ceil(value) // 5 6 6
Math.round(value) // 5 5 6
Math.trunc(value) // 5 5 5
parseInt(value) // 5 5 5
~~value // 5 5 5
value | 0 // 5 5 5
value >> 0 // 5 5 5
value >>> 0 // 5 5 5
value - value % 1 // 5 5 5
Negative
// value=x // x=-5 -5>x>=-5.5 -5.5>x>-6
Math.floor(value) // -5 -6 -6
Math.ceil(value) // -5 -5 -5
Math.round(value) // -5 -5 -6
Math.trunc(value) // -5 -5 -5
parseInt(value) // -5 -5 -5
value | 0 // -5 -5 -5
~~value // -5 -5 -5
value >> 0 // -5 -5 -5
value >>> 0 // 4294967291 4294967291 4294967291
value - value % 1 // -5 -5 -5
Positive - Larger numbers
// x = Number.MAX_SAFE_INTEGER/10 // =900719925474099.1
// value=x x=900719925474099 x=900719925474099.4 x=900719925474099.5
Math.floor(value) // 900719925474099 900719925474099 900719925474099
Math.ceil(value) // 900719925474099 900719925474100 900719925474100
Math.round(value) // 900719925474099 900719925474099 900719925474100
Math.trunc(value) // 900719925474099 900719925474099 900719925474099
parseInt(value) // 900719925474099 900719925474099 900719925474099
value | 0 // 858993459 858993459 858993459
~~value // 858993459 858993459 858993459
value >> 0 // 858993459 858993459 858993459
value >>> 0 // 858993459 858993459 858993459
value - value % 1 // 900719925474099 900719925474099 900719925474099
Negative - Larger numbers
// x = Number.MAX_SAFE_INTEGER/10 * -1 // -900719925474099.1
// value = x // x=-900719925474099 x=-900719925474099.5 x=-900719925474099.6
Math.floor(value) // -900719925474099 -900719925474100 -900719925474100
Math.ceil(value) // -900719925474099 -900719925474099 -900719925474099
Math.round(value) // -900719925474099 -900719925474099 -900719925474100
Math.trunc(value) // -900719925474099 -900719925474099 -900719925474099
parseInt(value) // -900719925474099 -900719925474099 -900719925474099
value | 0 // -858993459 -858993459 -858993459
~~value // -858993459 -858993459 -858993459
value >> 0 // -858993459 -858993459 -858993459
value >>> 0 // 3435973837 3435973837 3435973837
value - value % 1 // -900719925474099 -900719925474099 -900719925474099
Bitwise OR operator
A bitwise or operator can be used to truncate floating point figures and it works for positives as well as negatives:
function float2int (value) {
return value | 0;
}
Results
float2int(3.1) == 3
float2int(-3.1) == -3
float2int(3.9) == 3
float2int(-3.9) == -3
Performance comparison?
I've created a JSPerf test that compares performance between:
Math.floor(val)val | 0bitwise OR~~valbitwise NOTparseInt(val)
that only works with positive numbers. In this case you're safe to use bitwise operations well as Math.floor function.
But if you need your code to work with positives as well as negatives, then a bitwise operation is the fastest (OR being the preferred one). This other JSPerf test compares the same where it's pretty obvious that because of the additional sign checking Math is now the slowest of the four.
Note
As stated in comments, BITWISE operators operate on signed 32bit integers, therefore large numbers will be converted, example:
1234567890 | 0 => 1234567890
12345678901 | 0 => -539222987
There are functions to round numbers. For example:
var x = 5.0364342423;
print(x.toFixed(2));
will print 5.04.
EDIT: Fiddle
var result = Math.round(original*100)/100;
The specifics, in case the code isn't self-explanatory.
edit: ...or just use toFixed, as proposed by Tim Büthe. Forgot that one, thanks (and an upvote) for reminder :)
From the Floating-Point Guide:
What can I do to avoid this problem?
That depends on what kind of calculations you’re doing.
- If you really need your results to add up exactly, especially when you work with money: use a special decimal datatype.
- If you just don’t want to see all those extra decimal places: simply format your result rounded to a fixed number of decimal places when displaying it.
- If you have no decimal datatype available, an alternative is to work with integers, e.g. do money calculations entirely in cents. But this is more work and has some drawbacks.
Note that the first point only applies if you really need specific precise decimal behaviour. Most people don't need that, they're just irritated that their programs don't work correctly with numbers like 1/10 without realizing that they wouldn't even blink at the same error if it occurred with 1/3.
If the first point really applies to you, use BigDecimal for JavaScript or DecimalJS, which actually solves the problem rather than providing an imperfect workaround.
I like Pedro Ladaria's solution and use something similar.
function strip(number) {
return (parseFloat(number).toPrecision(12));
}
Unlike Pedros solution this will round up 0.999...repeating and is accurate to plus/minus one on the least significant digit.
Note: When dealing with 32 or 64 bit floats, you should use toPrecision(7) and toPrecision(15) for best results. See this question for info as to why.