Why does the following evaluate to 'hi'?
'hi' || true || 50
I'm not super new to javascript, but I'm rebeefing my knowledge by going through some old books and I for the life of me do not understand why this evaluates to 'hi' instead of true.. Can someone explain this??
Welcome to the world of truthy and falsey values.
If a value can be converted to true, the value is so-called truthy. If
a value can be converted to false, the value is so-called falsy.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Logical_Operators
This means that basically everything except
false
null
undefined
NaN
""
0
Will evaluate to true in || conditions, returning the first value that is truthy. This is sometimes used in a coalesce-like way:
a = a || {}
Which will set a to a if a is none of the values above, else an empty javascript object.
Because 'hi' is a non-empty string literal which evaluates to true when treated as a boolean. The expression a || b || c returns the first expression which evaluates to true, in this case 'hi'.
From MDN (Logical Operators):
Returns expr1 if it can be converted to true; otherwise, returns expr2. Thus, when used with Boolean values, || returns true if either operand can be converted to true; if both can be converted to false, returns false.
Hey thanks everybody for your input. Yeah, now it makes sense because I remember how the first value that evaluates to true is the one that it will evaluate to. I guess I have to do some more studying on the truthy stuff because yeah, it is simple, but in a way it is somewhat confusing at times. Thanks again!!
Related
I know that in JavaScript you can do:
var oneOrTheOther = someOtherVar || "these are not the droids you are looking for...";
where the variable oneOrTheOther will take on the value of the first expression if it is not null, undefined, or false. In which case it gets assigned to the value of the second statement.
However, what does the variable oneOrTheOther get assigned to when we use the logical AND operator?
var oneOrTheOther = someOtherVar && "some string";
What would happen when someOtherVar is non-false?
What would happen when someOtherVar is false?
Just learning JavaScript and I'm curious as to what would happen with assignment in conjunction with the AND operator.
Basically, the Logical AND operator (&&), will return the value of the second operand if the first is truthy, and it will return the value of the first operand if it is by itself falsy, for example:
true && "foo"; // "foo"
NaN && "anything"; // NaN
0 && "anything"; // 0
Note that falsy values are those that coerce to false when used in boolean context, they are null, undefined, 0, NaN, an empty string, and of course false, anything else coerces to true.
&& is sometimes called a guard operator.
variable = indicator && value
it can be used to set the value only if the indicator is truthy.
Beginners Example
If you are trying to access "user.name" but then this happens:
Uncaught TypeError: Cannot read property 'name' of undefined
Fear not. You can use ES6 optional chaining on modern browsers today.
const username = user?.name;
See MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Optional_chaining
Here's some deeper explanations on guard operators that may prove useful in understanding.
Before optional chaining was introduced, you would solve this using the && operator in an assignment or often called the guard operator since it "guards" from the undefined error happening.
Here are some examples you may find odd but keep reading as it is explained later.
var user = undefined;
var username = user && user.username;
// no error, "username" assigned value of "user" which is undefined
user = { username: 'Johnny' };
username = user && user.username;
// no error, "username" assigned 'Johnny'
user = { };
username = user && user.username;
// no error, "username" assigned value of "username" which is undefined
Explanation: In the guard operation, each term is evaluated left-to-right one at a time. If a value evaluated is falsy, evaluation stops and that value is then assigned. If the last item is reached, it is then assigned whether or not it is falsy.
falsy means it is any one of these values undefined, false, 0, null, NaN, '' and truthy just means NOT falsy.
Bonus: The OR Operator
The other useful strange assignment that is in practical use is the OR operator which is typically used for plugins like so:
this.myWidget = this.myWidget || (function() {
// define widget
})();
which will only assign the code portion if "this.myWidget" is falsy. This is handy because you can declare code anywhere and multiple times not caring if its been assigned or not previously, knowing it will only be assigned once since people using a plugin might accidentally declare your script tag src multiple times.
Explanation: Each value is evaluated from left-to-right, one at a time. If a value is truthy, it stops evaluation and assigns that value, otherwise, keeps going, if the last item is reached, it is assigned regardless if it is falsy or not.
Extra Credit: Combining && and || in an assignment
You now have ultimate power and can do very strange things such as this very odd example of using it in a palindrome.
function palindrome(s,i) {
return (i=i || 0) < 0 || i >= s.length >> 1 || s[i] == s[s.length - 1 - i] && isPalindrome(s,++i);
}
In depth explanation here: Palindrome check in Javascript
Happy coding.
Quoting Douglas Crockford1:
The && operator produces the value of its first operand if the first operand is falsy. Otherwise it produces the value of the second operand.
1 Douglas Crockford: JavaScript: The Good Parts - Page 16
According to Annotated ECMAScript 5.1 section 11.11:
In case of the Logical OR operator(||),
expr1 || expr2 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; if both are false, returns
false.
In the given example,
var oneOrTheOther = someOtherVar || "these are not the droids you are looking for...move along";
The result would be the value of someOtherVar, if Boolean(someOtherVar) is true.(Please refer. Truthiness of an expression). If it is false the result would be "these are not the droids you are looking for...move along";
And In case of the Logical AND operator(&&),
Returns expr1 if it can be converted to false; otherwise, returns
expr2. Thus, when used with Boolean values, && returns true if both
operands are true; otherwise, returns false.
In the given example,
case 1: when Boolean(someOtherVar) is false: it returns the value of someOtherVar.
case 2: when Boolean(someOtherVar) is true: it returns "these are not the droids you are looking for...move along".
I see this differently then most answers, so I hope this helps someone.
To calculate an expression involving ||, you can stop evaluating the expression as soon as you find a term that is truthy. In that case, you have two pieces of knowledge, not just one:
Given the term that is truthy, the whole expression evaluates to true.
Knowing 1, you can terminate the evaluation and return the last evaluated term.
For instance, false || 5 || "hello" evaluates up until and including 5, which is truthy, so this expression evaluates to true and returns 5.
So the expression's value is what's used for an if-statement, but the last evaluated term is what is returned when assigning a variable.
Similarly, evaluating an expression with && involves terminating at the first term which is falsy. It then yields a value of false and it returns the last term which was evaluated. (Edit: actually, it returns the last evaluated term which wasn't falsy. If there are none of those, it returns the first.)
If you now read all examples in the above answers, everything makes perfect sense :)
(This is just my view on the matter, and my guess as to how this actually works. But it's unverified.)
I have been seeing && overused here at work for assignment statements. The concern is twofold:
1) The 'indicator' check is sometimes a function with overhead that developers don't account for.
2) It is easy for devs to just see it as a safety check and not consider they are assigning false to their var. I like them to have a type-safe attitude, so I have them change this:
var currentIndex = App.instance && App.instance.rightSideView.getFocusItemIndex();
to this:
var currentIndex = App.instance && App.instance.rightSideView.getFocusItemIndex() || 0;
so they get an integer as expected.
I am trying to understand a small piece of code
I thought !!in javascript does the not operator two times
var result = !!String("false");
so I thought for the above code it will return as false
but it returns true..and can you tell me why the type is boolean?
That is because,
String("false") will return "false". A string.
!"false" will be evaluated to false.
Since a non empty string will be considered as true
when it coerces to boolean.
!false will be evaluated to true.
String("false")
Resolves to:
"false"
Which is a string that is non-empty, so
!"false"
Resolves to
false
And finally
!false
Resolves to
true
I know that in JavaScript you can do:
var oneOrTheOther = someOtherVar || "these are not the droids you are looking for...";
where the variable oneOrTheOther will take on the value of the first expression if it is not null, undefined, or false. In which case it gets assigned to the value of the second statement.
However, what does the variable oneOrTheOther get assigned to when we use the logical AND operator?
var oneOrTheOther = someOtherVar && "some string";
What would happen when someOtherVar is non-false?
What would happen when someOtherVar is false?
Just learning JavaScript and I'm curious as to what would happen with assignment in conjunction with the AND operator.
Basically, the Logical AND operator (&&), will return the value of the second operand if the first is truthy, and it will return the value of the first operand if it is by itself falsy, for example:
true && "foo"; // "foo"
NaN && "anything"; // NaN
0 && "anything"; // 0
Note that falsy values are those that coerce to false when used in boolean context, they are null, undefined, 0, NaN, an empty string, and of course false, anything else coerces to true.
&& is sometimes called a guard operator.
variable = indicator && value
it can be used to set the value only if the indicator is truthy.
Beginners Example
If you are trying to access "user.name" but then this happens:
Uncaught TypeError: Cannot read property 'name' of undefined
Fear not. You can use ES6 optional chaining on modern browsers today.
const username = user?.name;
See MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Optional_chaining
Here's some deeper explanations on guard operators that may prove useful in understanding.
Before optional chaining was introduced, you would solve this using the && operator in an assignment or often called the guard operator since it "guards" from the undefined error happening.
Here are some examples you may find odd but keep reading as it is explained later.
var user = undefined;
var username = user && user.username;
// no error, "username" assigned value of "user" which is undefined
user = { username: 'Johnny' };
username = user && user.username;
// no error, "username" assigned 'Johnny'
user = { };
username = user && user.username;
// no error, "username" assigned value of "username" which is undefined
Explanation: In the guard operation, each term is evaluated left-to-right one at a time. If a value evaluated is falsy, evaluation stops and that value is then assigned. If the last item is reached, it is then assigned whether or not it is falsy.
falsy means it is any one of these values undefined, false, 0, null, NaN, '' and truthy just means NOT falsy.
Bonus: The OR Operator
The other useful strange assignment that is in practical use is the OR operator which is typically used for plugins like so:
this.myWidget = this.myWidget || (function() {
// define widget
})();
which will only assign the code portion if "this.myWidget" is falsy. This is handy because you can declare code anywhere and multiple times not caring if its been assigned or not previously, knowing it will only be assigned once since people using a plugin might accidentally declare your script tag src multiple times.
Explanation: Each value is evaluated from left-to-right, one at a time. If a value is truthy, it stops evaluation and assigns that value, otherwise, keeps going, if the last item is reached, it is assigned regardless if it is falsy or not.
Extra Credit: Combining && and || in an assignment
You now have ultimate power and can do very strange things such as this very odd example of using it in a palindrome.
function palindrome(s,i) {
return (i=i || 0) < 0 || i >= s.length >> 1 || s[i] == s[s.length - 1 - i] && isPalindrome(s,++i);
}
In depth explanation here: Palindrome check in Javascript
Happy coding.
Quoting Douglas Crockford1:
The && operator produces the value of its first operand if the first operand is falsy. Otherwise it produces the value of the second operand.
1 Douglas Crockford: JavaScript: The Good Parts - Page 16
According to Annotated ECMAScript 5.1 section 11.11:
In case of the Logical OR operator(||),
expr1 || expr2 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; if both are false, returns
false.
In the given example,
var oneOrTheOther = someOtherVar || "these are not the droids you are looking for...move along";
The result would be the value of someOtherVar, if Boolean(someOtherVar) is true.(Please refer. Truthiness of an expression). If it is false the result would be "these are not the droids you are looking for...move along";
And In case of the Logical AND operator(&&),
Returns expr1 if it can be converted to false; otherwise, returns
expr2. Thus, when used with Boolean values, && returns true if both
operands are true; otherwise, returns false.
In the given example,
case 1: when Boolean(someOtherVar) is false: it returns the value of someOtherVar.
case 2: when Boolean(someOtherVar) is true: it returns "these are not the droids you are looking for...move along".
I see this differently then most answers, so I hope this helps someone.
To calculate an expression involving ||, you can stop evaluating the expression as soon as you find a term that is truthy. In that case, you have two pieces of knowledge, not just one:
Given the term that is truthy, the whole expression evaluates to true.
Knowing 1, you can terminate the evaluation and return the last evaluated term.
For instance, false || 5 || "hello" evaluates up until and including 5, which is truthy, so this expression evaluates to true and returns 5.
So the expression's value is what's used for an if-statement, but the last evaluated term is what is returned when assigning a variable.
Similarly, evaluating an expression with && involves terminating at the first term which is falsy. It then yields a value of false and it returns the last term which was evaluated. (Edit: actually, it returns the last evaluated term which wasn't falsy. If there are none of those, it returns the first.)
If you now read all examples in the above answers, everything makes perfect sense :)
(This is just my view on the matter, and my guess as to how this actually works. But it's unverified.)
I have been seeing && overused here at work for assignment statements. The concern is twofold:
1) The 'indicator' check is sometimes a function with overhead that developers don't account for.
2) It is easy for devs to just see it as a safety check and not consider they are assigning false to their var. I like them to have a type-safe attitude, so I have them change this:
var currentIndex = App.instance && App.instance.rightSideView.getFocusItemIndex();
to this:
var currentIndex = App.instance && App.instance.rightSideView.getFocusItemIndex() || 0;
so they get an integer as expected.
This question already has answers here:
What is the !! (not not) operator in JavaScript?
(42 answers)
Closed 8 years ago.
I am by no means an expert at Javascript, but I have been reading Mark Pilgrim's "Dive into HTML5" webpage and he mentioned something that I would like a better understanding of.
He states:
Finally, you use the double-negative trick to force the result to a Boolean value (true or false).
function supports_canvas() {
return !!document.createElement('canvas').getContext;
}
If anyone can explain this a little better I would appreciate it!
A logical NOT operator ! converts a value to a boolean that is the opposite of its logical value.
The second ! converts the previous boolean result back to the boolean representation of its original logical value.
From these docs for the Logical NOT operator:
Returns false if its single operand can be converted to true; otherwise, returns true.
So if getContext gives you a "falsey" value, the !! will make it return the boolean value false. Otherwise it will return true.
The "falsey" values are:
false
NaN
undefined
null
"" (empty string)
0
Javascript has a confusing set of rules for what is considered "true" and "false" when placed in a context where a Boolean is expected. But the logical-NOT operator, !, always produces a proper Boolean value (one of the constants true and false). By chaining two of them, the idiom !!expression produces a proper Boolean with the same truthiness as the original expression.
Why would you bother? Because it makes functions like the one you show more predictable. If it didn't have the double negative in there, it might return undefined, a Function object, or something not entirely unlike a Function object. If the caller of this function does something weird with the return value, the overall code might misbehave ("weird" here means "anything but an operation that enforces Boolean context"). The double-negative idiom prevents this.
In javascript, using the "bang" operator (!) will return true if the given value is true, 1, not null, etc. It will return false if the value is undefined, null, 0, or an empty string.
So the bang operator will always return a boolean value, but it will represent the opposite value of what you began with. If you take the result of that operation and "bang" it again, you can reverse it again, but still end up with a boolean (and not undefined, null, etc).
Using the bang twice will take a value that could have been undefined, null, etc, and make it just plain false. It will take a value that could have been 1, "true", etc. and make it just plain true.
The code could have been written:
var context = document.createElement('canvas').getContext;
var contextDoesNotExist = !context;
var contextExists = !contextDoesNotExist;
return contextExists;
Using !!variable gives you a guarantee of typecast to boolean.
To give you a simple example:
"" == false (is true)
"" === false (is false)
!!"" == false (is true)
!!"" === false (is true)
But it doesn't make sense to use if you are doing something like:
var a = ""; // or a = null; or a = undefined ...
if(!!a){
...
The if will cast it to boolean so there is no need to make the implicit double negative cast.
! casts "something"/"anything" to a boolean.
!! gives the original boolean value back (and guarantees the expression is a boolean now, regardless to what is was before)
The first ! coerces the variable to a boolean type and inverts it. The second ! inverts it again (giving you the original (correct) boolean value for whatever you are checking).
For clarity you would be better off using
return Boolean(....);
document.createElement('canvas').getContext may evaluate to either undefined or an object reference. !undefined yields true, ![some_object] yields false. This is almost what we need, just inverted. So !! serves to convert undefined to false and an object reference to true.
It's to do with JavaScript's weak typing. document.createElement('canvas').getContext is a function object. By prepending a single ! it evaluates it as a boolean expression and flips the answer around. By prepending another !, it flips the answer back. The end result is that the function evaluates it as a boolean expression, but returns an actual boolean result rather than the function object itself. Prepending !! is a quick and dirty way to typecast an expression to a boolean type.
If document.createElement('canvas').getContext isn't undefined or null, it will return true. Otherwise it will return false.
Attacklab.wmd_env.buttons=Attacklab.wmd_env.buttons||_4;
what does the || do in this case?
Adds _4 to the array which is Attacklab.wmd_env.buttons?
The || operator in JavaScript returns the value on the left if that value does not evaluate to false, otherwise it returns the value on the right.
From Mozilla's Core JavaScript 1.5 Reference:
expr1 || expr2
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; if both are false, returns false.
So, in this case, if Attacklab.wmd_env.buttons doesn't have a value, it sets the value to _4.
It's a fancy way of writing
if(!Attacklab.wmd_env.buttons)
Attacklab.wmd_env.buttons = _4;
It's nice for providing default values. Keep in mind that not only null and undefined will trigger the conditional, but also 0, false and '', ie everything which is considered false in boolean contexts.
If Attacklab.wmd_env.buttons is null or undefined, it will be set to the default value _4.
The || operator checks whether the value provided on the left side of the expression is false (in a boolean context). If so it returns an alternate value indicated by the right side of the expression. Otherwise it returns the original value.
So for example the following code would set 'Foo' to a default value if it is null:
Foo = Foo || "Default Value"
This is sometimes called the Coalescing operator.
It is supported in other languages such as Ruby and Perl. C# has the ?? operator which does the same thing.