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In Javascript console.log( (NaN||9) ); // returns 9 , why is this so? [duplicate]

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JavaScript OR (||) variable assignment explanation
(12 answers)
Closed 5 years ago.
I've noticed when I have:
console.log(( NaN||9) ); // returns 9
Why is this so?
Why, when you evaluate - put parentheses around - NaN OR 9, it chooses the 9?
It works with 0, -3, "f". Can someone please tell me what is going on in the background to cause this outcome??

In JavaScript, NaN is a "falsy" value. The || operator evaluates to either its first argument (if that is "truthy") or else to the second argument (if the first argument is "falsy"). Hence NaN || 9 evaluates to 9.
From the docs for logical operators:
Logical operators are typically used with Boolean (logical) values. When they are, they return a Boolean value. However, the && and || operators actually return the value of one of the specified operands, so if these operators are used with non-Boolean values, they may return a non-Boolean value.
and for || specifically:
Returns expr1 if it can be converted to true; otherwise, returns expr2. Thus, when used with Boolean values, || returns true if either operand is true.

Short answer, because NaN is falsy.
Here's a list of falsy values in javascript.
The logical OR opertator || in javascript can be tricky, but once you know how it works its both straightforward and very handy. If it's left side is falsy, then it's result will be whatever it's right side is (in your case, the number 9).

Comparing NaN in javascript [duplicate]

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How do I test for NaN? [duplicate]
(2 answers)
Closed 7 years ago.
In Javascript, if we multiply a string with number we get NaN:
console.log("manas" * 5); // result is NaN
Why then does the following code result in false instead of true?
console.log("manas" * 5 == NaN) // results false

Use the isNaN function instead.
console.log(isNaN("manas" * 5));
http://www.w3schools.com/jsref/jsref_isnan.asp
NaN, not a number, is a special type value used to denote an unrepresentable value. With JavaScript, NaN can cause some confusion, starting from its typeof and all to the way the comparison is handled.
Several operations can lead to NaN as the result.
Because there are many ways to represent a NaN, it makes sense that one NaN will not be equal to another NaN.

NaN is a special numeric value which is not equal to anything including itself. To properly test for NaN you need to use isNaN function.
From specification:
Returns true if the argument coerces to NaN, and otherwise returns false.
Here is also useful note from the same ECMAScript section:
A reliable way for ECMAScript code to test if a value X is a NaN is an expression of the form X !== X. The result will be true if and only if X is a NaN.

NaN compares unequal (via ==, !=, ===, and !==) to any other value -- including to another NaN value. Use Number.isNaN() or isNaN() to most clearly determine whether a value is NaN. Or perform a self-comparison: NaN, and only NaN, will compare unequal to itself.
Testing against NaN

In Javascript,
NaN == NaN //always results false.
(self comparison of NaN to itself is always false in javascript)
refer:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/NaN
also see Is NaN equal to NaN?

Javascript is not that smart - when either side of the == operator is Nan, the whole thing evaluates as false (similar to how if leftside of || is false, it doesn't bother looking at right side)
So, even Nan == Nan will return false.
Here's a good article on Nan behaviour
http://ariya.ofilabs.com/2014/05/the-curious-case-of-javascript-nan.html

How does JavaScript evaluate if statement expressions?

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

Javascript if-statement vs. comparsion

Is it true that in if-statement JavaScript wrap the condition into a Boolean?
if(x) => if(Boolean(x))
Is it true that in comparison JavaScript wrap the compare elements into a Number?
a == b => Number(a) == Number(b)

Yes, and No.
For the first part, yes, that is essentially what the javascript does.
But for the latter, no. Not everything in JavaScript can be converted to a number. For example:
Number('abc') // => NaN
And Not-A-Numbers are not equal:
NaN == NaN // => false
So something like this:
Number('abc') == Number('abc') // => false!
But that's actually true with equality comparison.
'abc' == 'abc' // => true
As a side note, it's probably better to use === in JavaScript, which also checks the type of the values being compared:
0 == '0' // => true
0 === '0' // => false, because integer is not a string
More details about === can be read over here.

Yes, that's true, x is evaluated in a boolean context in this situation, so the equivalent of Boolean(x) is applied.
No, that's not true. It only looks that way because the coercitive equality operator == tries to convert a and b to the same type. Number() is only applied if either a or b is already a Number. For instance:
>>> 0x2A == 42
true // both 'a' and 'b' are numbers.
>>> "0x2A" == 42
true // 'a' is a string whose number coercion is equal to 'b'.
>>> "0x2A" == "42"
false // 'a' and 'b' are different strings.

Is it true that in if-statement JavaScript wrap the condition into a Boolean?
Usually yes.
Is it true that in comparison JavaScript wrap the compare elements into a Number?
Absolutely no.
Explanation
From JavaScript Language Specifications.
The if statement is defined at § 12.5 as:
if ( Expression ) Statement else Statement
It says that the Expression will be evaluated, converted with GetValue() and then tested after the ToBoolean() conversion.
Then the first assertion is true (but see later), the condition for the if statement is evaluated like is passed as parameter to the Boolean function. Please recall how JavaScript handles type conversion to boolean (§ 9.2):
undefined and null values are converted to false.
numbers are converted to false if ±0 or NaN otherwise they're converted to true.
strings are converted to false if empty otherwise always to true regardless their content.
objects are always converted to true.
Because of the call to GetValue() strictly speaking this assertion is not always true, take a look to § 8.7.1 where the standard describes how GetValue() works, here can happen some magic conversion before ToBoolean() is called.
The == operator is defined as in § 11.9.3.
As you can see it doesn't specify that operands must be (or will be treated as) numbers, the behavior of the operator is different and regulated by a series of rules based on the type of the operands. Then your second assertion is false. The case they're numbers (or one of them is a number) is just a special case in the algorithm, please note that at point 4 of the algorithm it says that if one of them is a number and the other one is a string then it'll be converted with ToNumber(), only in this case (with all the implications that this conversion has).
It's intuitive if you think that you can compare functions, strings or numbers, not every type can be converted to a numeric value.

Obfuscated javascript code with binary values?

This code outputs D. The question is HOW?
alert([][(![]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]][([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(![]+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[!+[]+!+[]]]()[([][(![]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]][([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(![]+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[!+[]+!+[]]]()+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+(![]+[])[+!+[]]](+[]+[+[]])[+!+[]]);
I understand that ![] is evaluated to false or 0 and so on, but how does it execute? And how can I convert this to something humans can understand and not only Jon Skeet?
Can someone break some piece of this code and explain me what's happening?

Well, evaluating the expression in parts, at the end it is equivalent to:
[]['sort']['call']()["btoa"]("00")[1]; // "D"
Which can be simplified to :
btoa("00")[1]; // "D"
How you can "decode it"?
Simply examine the operators used, for example, we can see at first that an array literal is used, then several bracket notation property accesses are done, and a couple of invocations.
How does it work?
The trick is to chain multiple type conversions, for example, to get the f letter:
(![]+[])[+[]]
If we examine the first part, in parentheses, ![]+[], we see a boolean negation, which will return false because an array object is always truthy, and then a concatenation.
That produces the string "false", then, the second part, we see a brackets applied to that string, to access a character, and the expression +[], which results in 0.
+[] gives zero because the Array's toString method returns an empty string, for an empty array like that one, and an empty string produces to zero when it is converted to number (the unary + operator in this example).
There are just tricks like that, things that produce a string, such "true", "false", "null", "undefined", etc... and more tricks to get the numerically.
For example to get a number -to access a character-, they use again cryptic type conversion:
+[]; // 0, equivalent to +""
+!+[]; // 1, equivalent to +true
!+[]+!+[]; // 2, equivalent to +true+true
!+[]+!+[]+!+[]; // 3, equivalent to +true+true+true

It does some tricks with the javascript type conversions. As CMS pointed out, it's equivalent to: []['sort']['call']()["btoa"]("00")[1];
They build the strings by pulling them out of things like false, etc.
For example, to get the s in "sort":
Get a "false": (![]+[]) -- ![] returns false and +[] converts it to a string.
Get the value 3 with: !+[]+!+[]+!+[] - each !+[] returns true, but when you add booleans you get an integer representation. e.g. true + true = 2
Get the s with string index access notation ("false"[3] = 's'): (![]+[]) [!+[]+!+[]+!+[]]
And now you have an s. They keep doing that until they have enough to access whichever method or property they want.

![] is false. ![] + [] evaluates to 'false'. +[] is 0 so !+[] is true, etc. javascript is very strange with its implied type conversions