Was just playing around with nodejs and chrome's console when I tested this:
[] == true // false
![] == true // false
!![] == true // true
How come? Isn't it wrong?
See the ECMAScript standard:
11.4.9 Logical NOT Operator ( ! )
The production UnaryExpression : ! UnaryExpression is evaluated as follows:
Let expr be the result of evaluating UnaryExpression.
Let oldValue be ToBoolean(GetValue(expr)).
If oldValue is true, return false.
Return true.
9.2 ToBoolean
The abstract operation ToBoolean converts its argument to a value of type Boolean according to Table 11:
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.
The result is false if the argument is the empty String (its length is zero);
otherwise the result is true.
Object → true
An array is an Object.
It has to do with how the browser handles implicit conversions in JS.
As [] is an empty array, it evaluates to false:
[] == true
false == true
However, noting it, will turn the object into a boolean with a value of false:
![] == true
false == true
However, boolean([]) will return true.
However, noting that, will turn it into a boolean with the value of !false:
!![] == true
!false == true
true == true
This is why implicit conversions arn't recomended unless needed:
"1" == true
true == true
"1" === true
false
[] == true is false because [] is not equal to true, just like "some string" is also not equal to true.
![] == true is false because [] evaluates to a true value when used in a conditional statement:
if([]) console.log('[]');
if(![]) console.log('![]');
// the result will be '[]' because [] will evaluate to true
// in a conditional even though it doesn't equal true
Another thing that may help you to think about it is ![] == false will be true.
!![] == true is true because !! converts anything to a true or false value based on whether it would be true or false in a conditional statement. So if(obj) and if(!!obj) will always have the same result.
Related
I have two statements like this. Why do they both evaluate to false?
console.log([] == true)
console.log(![] == true)
If [] == true is false shouldn't ![] == true result to true?
It's the way coercion works.
The first step of coercion is to convert any non primitive types to primitive types, then using a set of rules convert the left, right or both sides to the same type. You can find these rules here.
In your case [] == true, would pass through these 4 steps:
[] == true
[] == 1
"" == 1
0 == 1
Whereas based on operator precedence the ! in ![] == true is executed first so the expression is converted to false == true which is obviously false.
You can try the live demo by Felix Kling to better understand how the sameness operator works.
In more words:
The value ![] is false, because [] is an Object (arrays are objects) and all objects, not including null, are truthy. So any array, even if it is empty will always be a truthy, and the opposite of a truthy is always false. The easiest way to check if a value is a truthy is by using !!.
console.log("![]: " + ![]);
console.log("!![]: " + !![]);
When you are using a loose comparison (using == instead of ===) you are asking the JavaScript engine to first convert the values into the same type and then compare them. So what happens is the values get converted according to a set of rules, once they are the same type, they get compared though the same process as a strict equality check (===). The first change that [] goes through is to get converted to a string, the string version of an empty array is an empty string "", the empty string is then converted to a number, the numeric value of an empty string is 0, since the numeric value of true is 1 and 0 != 1, the final output is false.
console.log("[] == true => `" + ([] == true) + "`");
console.log("String([]) => `" + String([]) + "`");
console.log("Number('') => `" + Number("") + "`");
console.log("Number(true) => `" + Number(true) + "`");
As per the Abstract Equality Comparison Algorithm - http://es5.github.io/#x11.9.3
Types of x and y are checked when x == y is to be checked.
If no rule matches, a false is returned.
For [] == true , rule 7 matches, so a result of [] == ToNumber(true) is returned i.e. false is returned.
Reason you're getting the same result for ![] == true, because ![] returns false, and false == true returns false .
To get opposite result for your second operation, add a precedence (i.e. wrap) to your expression with braces.
console.log(!([] == true)); // returns true
Put ![] in the console. It's false.
So ![] == true is the same as false == true, which is false.
[] != true is true though
None of the answers so far has addressed the main problem. [Edit: This is no longer true. See Nick Zoum's answer.]
The question essentially is this:
[] == true returns false
=> [] is not true
=> [] is false (because something not true must be false)
=> ![] should be true, since negation of false is true. Then why does ![] == true return false?
[] is actually truthy. (So ![] is actually falsy.)
Still, [] == true returns false -- the reason is coercion. It's another of Javascript's gotchas.
When an array is checked for equality explicitly against a boolean value (that is, against true or false), its elements are first converted to strings and then joined together. An empty array therefore becomes "" — therein lies the problem, because an empty string is falsy.
In brief, an empty array is truthy, but it's coerced (when being compared to a boolean value) to an empty string, which is falsy.
Note the following, which comes from https://www.nfriedly.com/techblog/2009/07/advanced-javascript-operators-and-truthy-falsy/
Arrays are particularly weird. If you just test it for truthyness, an
empty array is truthy. HOWEVER, if you compare an empty array to a
boolean, it becomes falsy:
if ( [] == false ) {
// this code runs }
if ( [] ) {
// this code also runs }
if ([] == true) {
// this code doesn't run }
if ( ![] ) {
// this code also doesn't run }
(This is because when you do a comparison, its elements are turned to Strings and joined. Since it's empty, it becomes "" which is then falsy. Yea, it's weird.)
Read https://javascriptweblog.wordpress.com/2011/02/07/truth-equality-and-javascript/
Rule in JavaScript for The Equals Operator (==)
if
type(x) type(y) result
1.x and y are the same type See Strict Equality (===) Algorithm
2.null Undefined true
3.Undefined null true
4. Number String x == toNumber(y)
5. String Number toNumber(x) == y
6. Boolean (any) toNumber(x) == y
7. (any) Boolean x == toNumber(y)
8.String or Number Object x == toPrimitive(y)
9.Object String or Number toPrimitive(x) == y
***10.otherwise… false
I have two statements like this. Why do they both evaluate to false?
console.log([] == true)
console.log(![] == true)
If [] == true is false shouldn't ![] == true result to true?
It's the way coercion works.
The first step of coercion is to convert any non primitive types to primitive types, then using a set of rules convert the left, right or both sides to the same type. You can find these rules here.
In your case [] == true, would pass through these 4 steps:
[] == true
[] == 1
"" == 1
0 == 1
Whereas based on operator precedence the ! in ![] == true is executed first so the expression is converted to false == true which is obviously false.
You can try the live demo by Felix Kling to better understand how the sameness operator works.
In more words:
The value ![] is false, because [] is an Object (arrays are objects) and all objects, not including null, are truthy. So any array, even if it is empty will always be a truthy, and the opposite of a truthy is always false. The easiest way to check if a value is a truthy is by using !!.
console.log("![]: " + ![]);
console.log("!![]: " + !![]);
When you are using a loose comparison (using == instead of ===) you are asking the JavaScript engine to first convert the values into the same type and then compare them. So what happens is the values get converted according to a set of rules, once they are the same type, they get compared though the same process as a strict equality check (===). The first change that [] goes through is to get converted to a string, the string version of an empty array is an empty string "", the empty string is then converted to a number, the numeric value of an empty string is 0, since the numeric value of true is 1 and 0 != 1, the final output is false.
console.log("[] == true => `" + ([] == true) + "`");
console.log("String([]) => `" + String([]) + "`");
console.log("Number('') => `" + Number("") + "`");
console.log("Number(true) => `" + Number(true) + "`");
As per the Abstract Equality Comparison Algorithm - http://es5.github.io/#x11.9.3
Types of x and y are checked when x == y is to be checked.
If no rule matches, a false is returned.
For [] == true , rule 7 matches, so a result of [] == ToNumber(true) is returned i.e. false is returned.
Reason you're getting the same result for ![] == true, because ![] returns false, and false == true returns false .
To get opposite result for your second operation, add a precedence (i.e. wrap) to your expression with braces.
console.log(!([] == true)); // returns true
Put ![] in the console. It's false.
So ![] == true is the same as false == true, which is false.
[] != true is true though
None of the answers so far has addressed the main problem. [Edit: This is no longer true. See Nick Zoum's answer.]
The question essentially is this:
[] == true returns false
=> [] is not true
=> [] is false (because something not true must be false)
=> ![] should be true, since negation of false is true. Then why does ![] == true return false?
[] is actually truthy. (So ![] is actually falsy.)
Still, [] == true returns false -- the reason is coercion. It's another of Javascript's gotchas.
When an array is checked for equality explicitly against a boolean value (that is, against true or false), its elements are first converted to strings and then joined together. An empty array therefore becomes "" — therein lies the problem, because an empty string is falsy.
In brief, an empty array is truthy, but it's coerced (when being compared to a boolean value) to an empty string, which is falsy.
Note the following, which comes from https://www.nfriedly.com/techblog/2009/07/advanced-javascript-operators-and-truthy-falsy/
Arrays are particularly weird. If you just test it for truthyness, an
empty array is truthy. HOWEVER, if you compare an empty array to a
boolean, it becomes falsy:
if ( [] == false ) {
// this code runs }
if ( [] ) {
// this code also runs }
if ([] == true) {
// this code doesn't run }
if ( ![] ) {
// this code also doesn't run }
(This is because when you do a comparison, its elements are turned to Strings and joined. Since it's empty, it becomes "" which is then falsy. Yea, it's weird.)
Read https://javascriptweblog.wordpress.com/2011/02/07/truth-equality-and-javascript/
Rule in JavaScript for The Equals Operator (==)
if
type(x) type(y) result
1.x and y are the same type See Strict Equality (===) Algorithm
2.null Undefined true
3.Undefined null true
4. Number String x == toNumber(y)
5. String Number toNumber(x) == y
6. Boolean (any) toNumber(x) == y
7. (any) Boolean x == toNumber(y)
8.String or Number Object x == toPrimitive(y)
9.Object String or Number toPrimitive(x) == y
***10.otherwise… false
Can someone explain me how this function works? I know the ternary operator, so I understand the part, that he checks if flag is true, and if not then use the 2. value. But I don't understand why flag is a Boolean in the first place and why it switches from true to false and vice versa when animate() is being called...
flag = $texte.data('animToggle');
$texte.data('animToggle', !flag);
$texte.animate({
'left': flag ? '-51%' : '0'
})
JavaScript has truthy and falsy values. These are values which equate to true and false when evaluated as Boolean. This means that true and false aren't the only values which can represent true or false data.
For example, if $texte.data('animToggle') is equal to 0 it will be falsy as 0 is a falsy value.
Example 1: 1000
1000 is a truthy value, so this will output "True":
var flag = 1000;
console.log(flag ? "True" : "False");
> "True"
Example 2: ""
"" (empty string) is a falsy value, so this will output "False":
var flag = "";
console.log(flag ? "True" : "False");
> "False"
The ! (Logical NOT) operator converts a value to Boolean and negates it. $texte.data('animToggle', !flag); sets the animToggle data property of $texte to the opposite of what it previously was.
!true // False
!1000 // False
!false // True
!"" // True
To the best of my knowledge, (x == false) should do the same thing as !x, as both of them try to interpret x as a boolean, and then negates it.
However, when I tried to test this out, I started getting some extremely strange behavior.
For example:
false == [] and false == ![] both return true.
Additionally
false == undefined and true == undefined both return false, as does
false == Infinity and true == Infinity and
false == NaN and true == NaN.
What exactly is going on here?
http://jsfiddle.net/AA6np/1/
It's all here: http://es5.github.com/#x11.9.3
For the case of false == []:
false is converted to a number (0), because that is always done with booleans.
[] is converted to a primitive by calling [].valueOf().toString(), and that is an empty string.
0 == "" is then evaluated by converting the empty string to a number, and because the result of that is also 0, false == [] is true.
For the case of false == ![]:
The logical not operator ! is performed by returning the opposite of ToBoolean(GetValue(expr))
ToBoolean() is always true for any object, so ![] evaluates to false (because !true = false), and therefore is false == ![] also true.
(false == undefined) === false and (true == undefined) === false is even simpler:
false and true are again converted to numbers (0 and 1, respectively).
Because undefined cannot be compared to a number, the chain bubbles through to the default result and that is false.
The other two cases are evaluated the same way: First Boolean to Number, and then compare this to the other number. Since neither 0 nor 1 equals Infinity or is Not A Number, those expressions also evaluate to false.
The abstract equality algorithm is described in section 9.3 of the specification.
For x == y where x = false and y = []:
Nope. Types are not equal.
Nope, x is not null.
Nope. x is not undefined.
Nope, x is not a number
Nope, x is not a string.
Yes, x is a boolean, so we compare ToNumber(x) and y.
Repeat the algorithm, x=0 and y=[].
We end at step 8:Type(x) == number. and Type(y) == object.
So, let the result be x == ToPrimitive(y).
ToPrimitive([]) == ""
Now, repeat the algorithm again with x=0 and y="". We end at 4: "return the result of the comparison x == ToNumber(y)."
ToNumber("") == 0
The last repetition of the algorithm ends at step 1 (types are equal). By 1.c.iii, 0 == 0, and true is returned.
The other results can be obtained in a similar manner, by using the algorithm.
false == []
Using == Javascript is allowed to apply conversions. The object will convert into a primitive to match type with the boolean, leaving an empty string. The false will convert into a number 0. The compares the empty string and the number 0. The string is converted to a number which will be 0, so the expression is "true"
![]
Javascript converts the object to the boolean true, therefore denying true ends being false.
false == undefined true == undefined
false == Infinity and true == Infinity
false == NaN and true == NaN
Again a bit of the same! false is converted to 0, true to 1. And then, undefined is converted to a number which is... NaN! So false in any case
I would recommend to use === !== as much as you can to get "expected" results unless you know very well what you are doing. Using something like Boolean(undefined) == false would also be nice.
Check ECMAScript specifications for all the details when converting stuff.
Many of us have seen / use the following:
var t=!0; (t will equal true)
var t=!1; (t will equal false)
How safe is this method of evaluating true / false?
I see google use this all the time.
Will these ALWAYS return a boolean true / false?
Thanks.
Yes:
11.4.9 Logical NOT Operator ( ! )
The production UnaryExpression : ! UnaryExpression is evaluated as
follows:
Let expr be the result of evaluating UnaryExpression.
Let oldValue be ToBoolean(GetValue(expr)).
If oldValue is true, return false.
Return true.
Source: http://es5.github.com/x11.html#x11.4.9
Yes, they'll always return true and false.
More specifically, !x will return true for any "falsey" value (undefined, "", etc) and conversely return false for any "truthy" value.
A more common idiom is !!x which converts any "falsey" value to strictly false and any "truthy" value to strictly true - in other words it's like a "type cast" to a boolean.