Develop Reference

Conditional `&&` operator on object property renders the value of the property in React [duplicate]

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.

Why the Javascript logical operator behaving this way? [duplicate]

I am unable to understand this.
Following is expression uses OR operator
var subCond1 = adj.getData('relationEnabled') == 'true' || adj.getData('unhideIfHidden') || adj.getData('hlFixed') == 'true';
I am expecting that since it is OR operation, it should return boolean true/false, but instead I get string 'false' as a result.
Can anyone explain this please?

Yup, that's just one of the features of || in JavaScript, and it's deliberate. It doesn't return a boolean (necessarily), it works like this: It evaluates the left-hand operand and if that operand is truthy, it returns it; otherwise, it evalutes and returns the right-hand operand.
So what's "truthy"? Anything that isn't "falsy". :-) The falsy values are 0, "", null, undefined, NaN, and of course, false. Anything else is truthy.
If you need a boolean, just !! the result:
var subCond1 = !!(adj.getData('relationEnabled') == 'true' || adj.getData('unhideIfHidden') || adj.getData('hlFixed') == 'true');
...but you frequently don't need to bother.
This behavior of || is really useful, particularly (I find) when dealing with object references that may be null, when you want to provide a default:
var obj = thisMayBeNull || {};
Now, obj will be thisMayBeNull if it's truthy (non-null object references are truthy), or {} if thisMayBeNull is falsy.
More in this article on my blog: JavaScript's Curiously-Powerful OR Operator (||)
Just to round things out: The && operator has a similar behavior: It evaluates the left-hand operand and, if it's falsy, returns it; otherwise it evaluates and returns the right-hand operator. This is useful if you want an object property from a variable which may be null:
var value = obj && obj.property;
value will be the value of obj if obj is falsy (for instance, null), or the value of obj.property if obj is truthy.

Javascript returns the first operand that has a truthy value, whatever that truthy value is or the value of the last operand if none before are truthy. That is a designed feature of Javascript (yes it is different than other languages).
You can turn that into a boolean by comparing to see if it is == true if you want or it is sometimes done with !!.

String comparison using "&&" [duplicate]

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.

Javascript && operator - clarification? [duplicate]

This question already has answers here:
Why don't logical operators (&& and ||) always return a boolean result?
(9 answers)
Closed 8 years ago.
edit :
Quote removed. is not related to the question.
Question
How come does it always (when all are truthy) takes the last value ?
Examples :
f= 1 && 2 && 3 //3
f= 'k' && 'd' && 's' //'s'
p.s. : I've noticed that when investigating the pub/sub using $.callbacks

Operator && acts as an if-statement. The short-circuit occurs only if a value is false. Otherwise, it keeps evaluating statements.
When you write a && b it means: a ? b : a.
Similarly, a || b means: a ? a : b.

Edit: author has clarified the question significantly.
JavaScript logical AND operator return the left operand if it is falsy or else it returns the right operand.
Let us take a look at the table of logical truth:
A B A && B
true true true
true false false
false true false
false false false
As you can see, if A is false, we can return it:
A A && B
false false
false false
If A is not false, we can return B:
B A && B
true true
false false
This is a minor optimization, which leads to interesting sideeffects when used with weak typing, such as inline if:
var result = callback && callback()
// Generally equivalent to a simple if:
if (callback)
result = callback();
else
result = null;
Also using default options:
name = name || 'John';
// Generally equivalent to a simple if:
if (!name)
name = 'John';
As to why it happens, well, because these operators are defined like that in ECMA-262.

So how come does it always (when all are truthy) takes the last value ?
Because it can't short-circuit if the leading values are truthy, because it's an "AND" operator, and all operands must be truthy to satisfy the condition.
Re your comment below:
yes but why does it returns the last value ?
Because it's useful. :-) In the very first version of JavaScript from Netscape, it just returned true if the overall expression was true and false if not, but almost immediately they realized that it's much more useful if it returns the final operand.
For instance, you can do this:
var value = obj && obj.value;
...which will set value to be either a falsey value (if obj is falsey) or obj.value (if obj is not falsey). Importantly, it won't throw an error if obj is falsey. Very handy for guarding if you don't know for sure that obj is set. You don't have to stop at one level, either:
var value = obj && obj.sub && obj.sub.value;
value will either be falsey, or the value of obj.sub.value if both obj and obj.sub are truthy (e.g., because we seem to be using them as objects, they're object instances rather than null or undefined).
|| does the same sort of thing: It returns its first truthy operand.
More: JavaScript's Curiously Powerful OR Operator (||)

Because if first one is truthy, then it should go on and check second operand till the last one. If you would go with 1 || 2 || 3 instead it will return 1.

&& is a logical operator. This tests whether a value is truthy or falsy. This linked article explains that falsy values are equal to undefined, null, NaN, 0, "" and false.
Logical expressions are checked from left to right. If any part of the expression is evaluated t be falsy, the remainder of the expression will not be evaluated. In your examples, the value is always truthy. Here we can break down your first expression:
f = 1 && 2 && 3;
f = 1 /* Okay, 1 is a truthy value, lets continue... */
f = 2 /* Still truthy, let's continue... */
f = 3 /* Still truthy, we've reached the end of the expression, f = 3. */
If any of the values before 3 were falsy, the expression would have ended. To show this in action, simply redeclare your variable as:
f = 1 && 2 && 0 && 3;
Here 0 is a falsy value (as mentioned above). The execution of your expression here will end at 0:
f = 1 /* Okay, 1 is a truthy value, lets continue... */
f = 2 /* Still truthy, let's continue... */
f = 0 /* Hey, this is falsy, lets stop here, f = 0. */
f = 3 /* We never get here as the expression has ended */
Here f ends up being 0.
In your jQuery.Topic() example, && is used to ensure that id actually exists:
topic = id && topics[id]
We can break this down in the same way:
/* Assume this is your topics array: */
topics = ["", "My First Topic", "My Second Topic"];
/* If we pass in the value 1 as "jQuery.Topic(1)": */
topic = id && topics[id]
topic = 1 /* This is truthy, let's continue... */
topic = topics[1] /* topics[1] === "My First Topic" */
/* Now lets assume we pass in nothing as "jQuery.Topic()": */
topic = id && topics[id]
topic = null /* This is falsy, let's stop here, topic = null. */
The following if statement will only work if topic is truthy. In the first example, this is the case as topic has a value which is set to a string which isn't empty. In the second example, topic is null which is falsy.

The ECMAScript Standard states 12.11.2 Runtime Semantics: Evaluation:
LogicalANDExpression : LogicalANDExpression && BitwiseORExpression
1. Let lref be the result of evaluating LogicalANDExpression.
2. Let lval be GetValue(lref).
3. Let lbool be ToBoolean(lval).
4. ReturnIfAbrupt(lbool).
5. If lbool is false, return lval.
6. Let rref be the result of evaluating BitwiseORExpression.
7. Return GetValue(rref).
That is, first we check whether the first operand is evaluated to be a "falsy" value (null, false, 0). If it is, then the first operand is returned.
If, however, it's "thruthy", then the same process is repeated for the second operand and if it is evaluated as a truthy value, then the original value is returned (7. Return GetValue(rref)).

Short-circuit evaluation and returning the second expression instead of a bool are quite useful features.
The && operator evaluates the left operand and if it's truthy evaluates the right operand. Similarly the || operator evaluates the left operand and if it's falsy evaluates the right operand.
This allows to write things like
draw_element(color || "#F00");
with the meaning of using color unless if it's not specified (e.g. color is null, undefined or "") and in that case the default value is used.
You can even use things like
draw_element(color || pick_new_color());
and the function pick_new_color() will be called only if a color is not specified.
A very common use of && instead is to avoid errors, e.g:
if (x && x.constructor == Array) {
...
}
because if x is null or undefined you cannot access attributes (an exception would be raised if you try). The fact that && returns the first result if it's falsy instead of just false is not so useful indeed. Your code example does it, but it's not something you will find often in Javascript programs.
To recap: && and || are short-circuiting (i.e. they don't even evaluate the second operand if the result is known after evaluating the first) and this is very useful. They also return the falsy/truthy value itself instead of a bool, and this is very useful for || (not so much useful for &&).

Javascript AND operator within assignment

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.