Related
Soooooo isNaN is apparently broken in JavaScript, with things like:
isNaN('')
isNaN(' ')
isNaN(true)
isNaN(false)
isNaN([0])
Returning false, when they appear to all be... Not a Number...
In ECMAScript 6, the draft includes a new Number.isNaN but it looks like (imo) that this is also broken...
I would expect
Number.isNaN('RAWRRR')
To return true, since it's a string, and cannot be converted to a number... However...
It seems that things that I would consider... not a number, are indeed, not, not a number...
http://people.mozilla.org/~jorendorff/es6-draft.html#sec-isfinite-number
The examples on MDN say:
Number.isNaN("blabla"); // e.g. this would have been true with isNaN
I don't understand how this is "More robust version of the original global isNaN." when I cannot check to see if things are not a number.
This would mean we're still subjected to doing actual type checking as well as checking isNaN... which seems silly...
http://people.mozilla.org/~jorendorff/es6-draft.html#sec-isnan-number
The ES3 draft here basically says, everything is always false, except with its Number.NaN
Does anyone else find this is broken or am I just not understanding the point of isNaN?
isNaN() and Number.isNaN() both test if a value is (or, in the case of isNaN(), can be converted to a number-type value that represents) the NaN value. In other words, "NaN" does not simply mean "this value is not a number", it specifically means "this value is a numeric Not-a-Number value according to IEEE-754".
The reason all your tests above return false is because all of the given values can be converted to a numeric value that is not NaN:
Number('') // 0
Number(' ') // 0
Number(true) // 1
Number(false) // 0
Number([0]) // 0
The reason isNaN() is "broken" is because, ostensibly, type conversions aren't supposed to happen when testing values. That is the issue Number.isNaN() is designed to address. In particular, Number.isNaN() will only attempt to compare a value to NaN if the value is a number-type value. Any other type will return false, even if they are literally "not a number", because the type of the value NaN is number. See the respective MDN docs for isNaN() and Number.isNaN().
If you simply want to determine whether or not a value is of the number type, even if that value is NaN, use typeof instead:
typeof 'RAWRRR' === 'number' // false
No, the original isNaN is broken. You are not understanding the point of isNaN.
The purpose of both of these functions is to determine whether or not something has the value NaN. This is provided because something === NaN will always be false and therefore can't be used to test this.
(side note: something !== something is actually a reliable, although counter-intuitive, test for NaN)
The reason isNaN is broken is that it can return true in cases when a value is not actually NaN. This is because it first coerces the value to a number.
So
isNaN("hello")
is true, even though "hello" is not NaN.
If you want to check whether a value actually is a finite number, you can use:
Number.isFinite(value)
If you want to test whether a value is a finite number or a string representation of one, you can use:
Number.isFinite(value) || (Number.isFinite(Number(value)) && typeof value === 'string')
As mentioned in a comment isNaN() and Number.isNaN() both check that the value you pass in is not equal to the value NaN. The key here is that NaN is an actual value and not an evaluated result e.g. "blabla" is a String and the value is "blabla" which means it is not the value "NaN".
A plausible solution would be doing something like:
Number.isNaN(Number("blabla")); //returns true.
The key difference between the two is that the global isNaN(x) function performs a conversion of the parameter x to a number. So
isNaN("blabla") === true
because Number("blabla") results in NaN
There are two definitions of "not a number" here and that's perhaps where the confusion lies. Number.isNaN(x) only returns true for the IEEE 754 floating point specification's definition of Not a Number, for example:
Number.isNaN(Math.sqrt(-1))
as opposed to determining whether the object being passed in is of numeric type or not. Some ways of doing that are:
typeof x === "number"
x === +x
Object.prototype.toString.call(x) === "[object Number]"
Basically, window.isNaN performs a type conversion to a number, then checks if it is NaN. Whereas, Number.isNaN doesn't try to convert its argument to a number. So basically, you can think of window.isNaN, and Number.isNaN as working like so.
window.isNaN = function(n){
return Number(n) !== Number(n);
}
window.Number.isNaN = function(n){
return n !== n;
}
Please note that you don't need actually to use the window. to call isNaN or Number.isNaN. Rather, I am just using it to provide a better distinction between the two similarly-named methods to try to cut down on confusion.
~ Happy Coding!
The following works because NaN is the only value in javascript which is not equal to itself.
Number.isNaN = Number.isNaN || function(value) {
return value !== value;
}
Per, MDN, it (NaN) is the returned value when Math functions fail and as such it is a specific value. Perhaps a better name would have been, MathFunctionFailed.
To determine if something is a number requires parsing which fails nicely over a broad range of non numeric inputs, successfully detecting numbers and strings representing numbers, hence:
function isNumber(v) {return Number.isNaN(parseFloat(v)); }
1. Number.isNaN
alert(Number.isNaN('Hello')); // false
Shouldn't it return true because Hello is a string and its Not A Number right ? But Lets know why it returns false.
MDN docs says :
true if the given value is NaN and its type is Number; otherwise,
false.
Yes Hello value is NaN but the type is string , you can check the type as follows:
alert(typeof `Hello`); // string
Usage:
Use when you want to check the value is both NaN and type is number.
2. isNaN
alert(isNaN('Hello')); // true
MDN docs says:
true if the given value is NaN; otherwise, false.
Usage:
Use when you want to check value is just NaN.
3. jQuery.isNumeric()
Jquery Docs Says :
Determines whether its argument represents a JavaScript number.
alert($.isNumeric('Hello')); // false
alert($.isNumeric(3)); //true
Usage:
Use when you want to check value is a number or can be converted to a number.
Reference
I use a simple workaround to check Number.isNaN():
let value = 'test';
Number.isNaN(-value); // true
value = 42;
Number.isNaN(-value); // false
js trying to convert the value to the negative Number, if the conversion is failed - we have NaN.
Simple, isn't it?
Moreover, online benchmark tests say Number.isNaN is lighter than isNaN.
Number.isNaN(x) checks if x is directly evaluated to NaN or not.
RAWR is not the same as NaN. Think of NaN as an entity to represent the result of some mathematical calculation where the computer does not know how to represent the number.
A mathematical operation is not going to yield a non-numeric result, hence the typeof NaN is number.
The string RAWR had undergone no mathematical operation to yield NaN. However, if you were to call Number.isNaN(+'RAWR'), it would result in NaN since the unary + operator is trying to convert 'RAWR' to a number.
On the other hand, isNaN(y) tells whether y can be converted to a number or not. If isNaN(y) is false, y can be converted to a number. But if isNaN(y) is true, y can not be converted to a number.
So a good rule of thumb is:
Do I want to check if x can be successfully converted to a number? Use isNaN(x) == false in that case. Unsuccessful conversion results in NaN.
Do I want to check if x is evaluated to NaN? Use Number.isNaN(x) == true for that.
#phill, as stated in other responses, neither is broken.
Number.isNaN work on a number or instance of Number, so even Number.isNaN(new Date(NaN)) is false.
isNaN, on the other hand, is generic and tries to convert its parameter to a number before checking it.
If you want to determine if a value is (or contains) NaN, you can use this function:
function valueIsNaN(value) {
// eslint-disable-next-line no-self-compare
return value !== value || typeof value == 'object' && Number.isNaN(value.valueOf());
}
I’ve heard of boolean arithmetic and thought of giving it a try.
alert (true+true===2) //true
alert (true-true===0) //true
So algebra tells me true=1
alert (true===1) //false :O
Could someone explain why this happens?
=== is the strict equality operator. Try == operator instead.
true==1 will evaluate to true.
The strict equality operator === only considers values equal if they
have the same type. The lenient equality operator == tries to
convert values of different types, before comparing like strict
equality.
Case 1:
In case of true===1, The data type of true is boolean whereas the type of 1 is number. Thus the expression true===1 will evaluate to false.
Case 2:
In case of true+true===2 and true-true===0 the arithmetic operation is performed first(Since + operator takes precedence over ===. See Operator Precedence) and then the result is compared with the other operand.
While evaluating expression (true+true===2), the arithmetic operation true+true performed first producing result 2. Then the result is compered with the other operand. i.e. (2==2) will evaluate to true.
Because comparing data TYPE and value (that's what operator '===' does ), TRUE is not exactly the same as 1. If you changed this to TRUE == 1, it should work fine.
At the beginning, you're doing bool + bool. The + operator takes precedence over the === operator so it's evaluated first. In this evaluation, it's converting the booleans to their number forms. Run console.log(true + true); and this will return 2. Since you're comparing the number 2 to the number 2, you get a return value true with the strict equality.
When you're just comparing true === 1, like everyone else said you're comparing the boolean true to the number 1 which is not strictly equal.
first 2 expression are true because you are using expression (true+true) (true-true) it convert type of a value first due to expression and check equality with "===", toNumber and toPrimitive are internal methods which convert their arguments (during expression ) this is how conversion take place during expression
That's why true+true equal to the 2
In your third expression you are using === this not convert arguments just check equality with type, to make it true both values and there type must be same.
Thats all
I always thought that JavaScript's if statements did some kind of casting magic to their arguments, but I'm a little wary of what's actually going on behind the scenes.
I recently found a JavaScript comparison table and noticed that even though -1 == true evaluates to false, if(-1){...} will execute.
So within JavaScripts if statements, what happens to the expression? It seems reasonable to assume that it uses !!{expression} to cast it to an inverse boolean, then invert it again, but if that's the case, how does JS decide whether an object's inverse boolean representation is truthy or not?
JavaScript is wonky.
Yes, -1 == true results in false, but that's not what the if statement is doing. It's checking to see if the statement is 'truthy', or converts to true. In JavaScript, that's the equivalent of !!-1, which does result in true (all numbers other than zero are truthy).
Why?!?
The spec defines the double equals operator to do the following when presented with a number and a boolean:
If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
ToNumber will convert the boolean true into the number 1, so you're comparing:
-1 == 1
which anyone can tell you is clearly false.
On the other hand, an if statement is calling ToBoolean, which considers any non-zero, non-NaN number to be true.
Any JavaScript developer really needs to look at the documentation -- for this case, located here: http://www.ecma-international.org/ecma-262/5.1/#sec-9.2
9.2 ToBoolean
The abstract operation ToBoolean converts its argument to a value of type Boolean according to Table 11:
Argument Type Result
Undefined false
Null false
Boolean The result equals the input argument (no conversion).
Number The result is false if the argument is +0, −0, or NaN; otherwise the result is true.
String The result is false if the argument is the empty String (its length is zero); otherwise the result is true.
Object true
(Sorry about the formatting, can't make a table here.)
From JavaScript The Definitive Guide
The following values convert to, and therefore work like, false:
undefined
null
0
-0
NaN
"" // the empty string
All other values, including all objects (and arrays) convert to, and work like, true. false, and the six values that convert to it, are sometimes called falsy values, and all other values are called truthy.
These things by themselves are falsy (or evaluate to false):
undefined
null
0
'' or ""
false
NaN
Everything else i truthy.
Truthy-ness or falsy-ness is used when evaluating a condition where the outcome is expected to be either truthy (true) or falsy (false).
In your example if(-1 == true), you are comparing apples and oranges. The compare is evaluated first (and resulted in false), and the results of that is used in your condition. The concept of truthyness/falsyness isn't applied to the operands the comparison.
When if state using with comparing variable different type js use .toString и .valueOf ( for more information check http://javascript.info/tutorial/object-conversion ) - just keep this in mind - it make so example much more easy to understand
So for what I know is that the if statement in Javascript casts the result of its condition to a Boolean, and then executes it likes the following
if(true) {
// run this
}
if(false) {
// do not run this
}
And that works. But If I do this:
if('0' == false) {
// We get here, so '0' is a falsy value
}
Then I would expect this
if('0') {
// We don't get here, because '0' is falsy value
}
But instead I get
if('0') {
// We *DO* get here, even though '0' is falsy value
}
So what's happening? Apparently, if does not check if its condition a truthy or falsy value, but does some other conversion?
This is just one of those "gotchas" with the == rules which are rather complex.
The comparison x == y, where x and y are values, produces true or false. Such a comparison is performed as follows:
(4) If Type(x) is Number and Type(y) is String,
return the result of the comparison x == ToNumber(y).
(5) If Type(x) is String and Type(y) is Number,
return the result of the comparison ToNumber(x) == y.
(6) If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.
(7) If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
In this case that means that '0' == false is first coerced do '0' == 0 (by rule #7) and then on the second pass through, it is coerced to 0 == 0 (by rule #5) which results in true.
This particular case is somewhat tricky because of false ~> 0 instead of '0' ~> true (as what might be expected). However, '0' is itself a truth-y value and the behavior can be explained with the above rules. To have strict truthy-falsey equality in the test (which is different than a strict-equality) without implicit conversions during the equality, consider:
!!'0' == !!false
(For all values: !falsey -> true and !truthy -> false.)
This:
if('0') {
// We *DO* get here, even though '0' is falsy value
}
checks to see if the string is null or empty, not whether it's zero or not. Any non-empty string is truthy.
When you do this:
if('0' == false) {
// We get here, so '0' is a falsy value
}
you are asking the JS engine for an explicit type conversion to try to match the type of the two operands. That is different than just asking if one operand is truthy all by itself.
In general, you will find that you get fewer unexpected results if you nearly always use === and !== and only allow type coercion when you know exactly what is going to happen in all cases either because you fully understand the very complex coercion rules or because you know what types will be present and you understand those specific cases.
if ('0' == false):
Javascript is doing something called type coercion.
Following the rules in that link, we fall to rule 7:
If Type(y) is Boolean, return the result of the comparison x == ToNumber(y)
Calling ToNumber(false) gives us a numeric 0. The result now starts to make sense, but we're still not quite done, because we still have a string and a number. The process starts again, and this time we fall to rule 5:
If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x) == y: "2" == 2
This time, the left side '0' is converted to a number: 0. Now, at last, we can compare two Numbers, and since 0 equals 0, the result is true. However, it's important to note that this implies nothing at all about the truish/falsy nature of the '0' string, because it was coerced before it was compared.
if('0')
In this case, there is no comparison; you only want to know if a single value is "truish" or "falsy". No type coercion is used, because strings can be evaluated as truish or falsy on their own merits. Using the rules at the same link as before, we find this information:
In JavaScript, and not only JavaScript, we have so called falsy values. These are respectively: 0, null, undefined, false, "", NaN. Please note the empty string is empty, 'cause differently from php as example, "0" will be considered truish
The quote is especially helpful because it specifically calls out the '0' string, but that would not be necessary. It's enough to know that an empty string is falsy, and any other string is truish, because the content of the string is not evaluated and no coercion is performed. 0 may be a falsy value, but because we evaluate a string rather than coercing to a number, and '0' has a value of some kind, it is still truish.
Javascript operator == does type conversion and is basically useless. Just avoid it.
For example:
[] is truty but [] == false is true
1 == "1", [1] == "1", [[1]] == "1" are all true
[1] == [[1]] however is false
The rules are VERY weird. For example in the first case [] gets converted to "" that gets converted to a number and the value is 0. false is also converted to the number 0. So in the end they compare equal.
Note however that while the conversion from the empty string to a number gives 0, the result of parseInt("") is NaN.
PS: The real fun is when you discover that [30,20,10,3,2,1].sort() returns [1,10,2,20,3,30] (yes... numbers, yes in lexicographical order). No I'm not kidding.
The following shows that "0" is false in Javascript:
>>> "0" == false
true
>>> false == "0"
true
So why does the following print "ha"?
>>> if ("0") console.log("ha")
ha
Tables displaying the issue:
and ==
Moral of the story use ===
table generation credit: https://github.com/dorey/JavaScript-Equality-Table
The reason is because when you explicitly do "0" == false, both sides are being converted to numbers, and then the comparison is performed.
When you do: if ("0") console.log("ha"), the string value is being tested. Any non-empty string is true, while an empty string is false.
Equal (==)
If the two operands are not of the same type, JavaScript converts the operands then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the other operand is converted to a string if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
(From Comparison Operators in Mozilla Developer Network)
It's according to spec.
12.5 The if Statement
.....
2. If ToBoolean(GetValue(exprRef)) is true, then
a. Return the result of evaluating the first Statement.
3. Else,
....
ToBoolean, according to the spec, is
The abstract operation ToBoolean converts its argument to a value of type Boolean according to Table 11:
And that table says this about strings:
The result is false if the argument is the empty String (its length is zero);
otherwise the result is true
Now, to explain why "0" == false you should read the equality operator, which states it gets its value from the abstract operation GetValue(lref) matches the same for the right-side.
Which describes this relevant part as:
if IsPropertyReference(V), then
a. If HasPrimitiveBase(V) is false, then let get be the [[Get]] internal method of base, otherwise let get
be the special [[Get]] internal method defined below.
b. Return the result of calling the get internal method using base as its this value, and passing
GetReferencedName(V) for the argument
Or in other words, a string has a primitive base, which calls back the internal get method and ends up looking false.
If you want to evaluate things using the GetValue operation use ==, if you want to evaluate using the ToBoolean, use === (also known as the "strict" equality operator)
It's PHP where the string "0" is falsy (false-when-used-in-boolean-context). In JavaScript, all non-empty strings are truthy.
The trick is that == against a boolean doesn't evaluate in a boolean context, it converts to number, and in the case of strings that's done by parsing as decimal. So you get Number 0 instead of the truthiness boolean true.
This is a really poor bit of language design and it's one of the reasons we try not to use the unfortunate == operator. Use === instead.
// I usually do this:
x = "0" ;
if (!!+x) console.log('I am true');
else console.log('I am false');
// Essentially converting string to integer and then boolean.
Your quotes around the 0 make it a string, which is evaluated as true.
Remove the quotes and it should work.
if (0) console.log("ha")
It is all because of the ECMA specs ... "0" == false because of the rules specified here http://ecma262-5.com/ELS5_HTML.htm#Section_11.9.3 ...And if ('0') evaluates to true because of the rules specified here http://ecma262-5.com/ELS5_HTML.htm#Section_12.5
== Equality operator evaluates the arguments after converting them to numbers.
So string zero "0" is converted to Number data type and boolean false is converted to Number 0.
So
"0" == false // true
Same applies to `
false == "0" //true
=== Strict equality check evaluates the arguments with the original data type
"0" === false // false, because "0" is a string and false is boolean
Same applies to
false === "0" // false
In
if("0") console.log("ha");
The String "0" is not comparing with any arguments, and string is a true value until or unless it is compared with any arguments.
It is exactly like
if(true) console.log("ha");
But
if (0) console.log("ha"); // empty console line, because 0 is false
`
This is because JavaScript uses type coercion in Boolean contexts and your code
if ("0")
will be coerced to true in boolean contexts.
There are other truthy values in Javascript which will be coerced to true in boolean contexts, and thus execute the if block are:-
if (true)
if ({})
if ([])
if (42)
if ("0")
if ("false")
if (new Date())
if (-42)
if (12n)
if (3.14)
if (-3.14)
if (Infinity)
if (-Infinity)
This is the reason why you should whenever possible use strict equality === or strict inequality !==
"100" == 100
true because this only checks value, not the data type
"100" === 100
false this checks value and data type
The "if" expression tests for truthiness, while the double-equal tests for type-independent equivalency. A string is always truthy, as others here have pointed out. If the double-equal were testing both of its operands for truthiness and then comparing the results, then you'd get the outcome you were intuitively assuming, i.e. ("0" == true) === true. As Doug Crockford says in his excellent JavaScript: the Good Parts, "the rules by which [== coerces the types of its operands] are complicated and unmemorable.... The lack of transitivity is alarming." It suffices to say that one of the operands is type-coerced to match the other, and that "0" ends up being interpreted as a numeric zero, which is in turn equivalent to false when coerced to boolean (or false is equivalent to zero when coerced to a number).
if (x)
coerces x using JavaScript's internal toBoolean (http://es5.github.com/#x9.2)
x == false
coerces both sides using internal toNumber coercion (http://es5.github.com/#x9.3) or toPrimitive for objects (http://es5.github.com/#x9.1)
For full details see http://javascriptweblog.wordpress.com/2011/02/07/truth-equality-and-javascript/
I have same issue, I found a working solution as below:
The reason is
if (0) means false, if (-1, or any other number than 0) means true. following value are not truthy, null, undefined, 0, ""empty string, false, NaN
never use number type like id as
if (id) {}
for id type with possible value 0, we can not use if (id) {}, because if (0) will means false, invalid, which we want it means valid as true id number.
So for id type, we must use following:
if ((Id !== undefined) && (Id !== null) && (Id !== "")){
} else {
}
for other string type, we can use if (string) {}, because null, undefined, empty string all will evaluate at false, which is correct.
if (string_type_variable) { }
In JS "==" sign does not check the type of variable. Therefore, "0" = 0 = false (in JS 0 = false) and will return true in this case, but if you use "===" the result will be false.
When you use "if", it will be "false" in the following case:
[0, false, '', null, undefined, NaN] // null = undefined, 0 = false
So
if("0") = if( ("0" !== 0) && ("0" !== false) && ("0" !== "") && ("0" !== null) && ("0" !== undefined) && ("0" !== NaN) )
= if(true && true && true && true && true && true)
= if(true)
I came here from search looking for a solution to evaluating "0" as a boolean.
The technicalities are explained above so I won't go into it but I found a quick type cast solves it.
So if anyone else like me is looking to evaluate a string 1 or 0 as boolean much like PHP. Then you could do some of the above or you could use parseInt() like:
x = "0";
if(parseInt(x))
//false