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Is there way to use the variable that tested true in an if statement

I can't explain it much in the title but I'm now stuck with having to test 2 variables for the same value and was wondering is there a way to do something like this.
var a = 'abc';
var b = 'def';
if (a=='abc' || b=='abc'){ c = 'z'; }
Now c is the variable that was tested true, so by assigning a value to c it would assign the value to the variable that tested true in the if statement. a and b can have different values but both can have the same value aswell.

No, the expression just yields a result value, with no information stored about which subexpression was truthy. (And separately, JavaScript doesn't have references to variables [at least, not that you can use in your code].) You'll have to break it out into two ifs:
if (a === "abc") {
a = "z";
} else if (b === "abc") {
b = "z";
}
It would be possible to do something like this if the variables were accessor properties (or other function calls) with side effects (since the right-hand operand of the || won't be evaluated at all if the left-hand operand evaluates to a truthy value). Or if they were properties in an object and you tested an array of property names, etc. But it would be unnecessarily complicated. :-)

You could take a switch statement, here you can check various variables agains a certain value.
switch ('abc') {
case a:
a = 'z';
break;
case b:
b = 'z';
break;
}

javascript object related query [duplicate]

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Is it ever possible that (a== 1 && a ==2 && a==3) could evaluate to true in JavaScript?
This is an interview question asked by a major tech company. It happened two weeks back, but I'm still trying to find the answer. I know we never write such code in our day-to-day job, but I'm curious.

If you take advantage of how == works, you could simply create an object with a custom toString (or valueOf) function that changes what it returns each time it is used such that it satisfies all three conditions.
const a = {
i: 1,
toString: function () {
return a.i++;
}
}
if(a == 1 && a == 2 && a == 3) {
console.log('Hello World!');
}
The reason this works is due to the use of the loose equality operator. When using loose equality, if one of the operands is of a different type than the other, the engine will attempt to convert one to the other. In the case of an object on the left and a number on the right, it will attempt to convert the object to a number by first calling valueOf if it is callable, and failing that, it will call toString. I used toString in this case simply because it's what came to mind, valueOf would make more sense. If I instead returned a string from toString, the engine would have then attempted to convert the string to a number giving us the same end result, though with a slightly longer path.

I couldn't resist - the other answers are undoubtedly true, but you really can't walk past the following code:
var aﾠ = 1;
var a = 2;
var ﾠa = 3;
if(aﾠ==1 && a== 2 &&ﾠa==3) {
console.log("Why hello there!")
}
Note the weird spacing in the if statement (that I copied from your question). It is the half-width Hangul (that's Korean for those not familiar) which is an Unicode space character that is not interpreted by ECMA script as a space character - this means that it is a valid character for an identifier. Therefore there are three completely different variables, one with the Hangul after the a, one with it before and the last one with just a. Replacing the space with _ for readability, the same code would look like this:
var a_ = 1;
var a = 2;
var _a = 3;
if(a_==1 && a== 2 &&_a==3) {
console.log("Why hello there!")
}
Check out the validation on Mathias' variable name validator. If that weird spacing was actually included in their question, I feel sure that it's a hint for this kind of answer.
Don't do this. Seriously.
Edit: It has come to my attention that (although not allowed to start a variable) the Zero-width joiner and Zero-width non-joiner characters are also permitted in variable names - see Obfuscating JavaScript with zero-width characters - pros and cons?.
This would look like the following:
var a= 1;
var a‍= 2; //one zero-width character
var a‍‍= 3; //two zero-width characters (or you can use the other one)
if(a==1&&a‍==2&&a‍‍==3) {
console.log("Why hello there!")
}

IT IS POSSIBLE!
var i = 0;
with({
get a() {
return ++i;
}
}) {
if (a == 1 && a == 2 && a == 3)
console.log("wohoo");
}
This uses a getter inside of a with statement to let a evaluate to three different values.
... this still does not mean this should be used in real code...
Even worse, this trick will also work with the use of ===.
var i = 0;
with({
get a() {
return ++i;
}
}) {
if (a !== a)
console.log("yep, this is printed.");
}

Example without getters or valueOf:
a = [1,2,3];
a.join = a.shift;
console.log(a == 1 && a == 2 && a == 3);
This works because == invokes toString which calls .join for Arrays.
Another solution, using Symbol.toPrimitive which is an ES6 equivalent of toString/valueOf:
let i = 0;
let a = { [Symbol.toPrimitive]: () => ++i };
console.log(a == 1 && a == 2 && a == 3);

If it is asked if it is possible (not MUST), it can ask "a" to return a random number. It would be true if it generates 1, 2, and 3 sequentially.
with({
get a() {
return Math.floor(Math.random()*4);
}
}){
for(var i=0;i<1000;i++){
if (a == 1 && a == 2 && a == 3){
console.log("after " + (i+1) + " trials, it becomes true finally!!!");
break;
}
}
}

When you can't do anything without regular expressions:
var a = {
r: /\d/g,
valueOf: function(){
return this.r.exec(123)[0]
}
}
if (a == 1 && a == 2 && a == 3) {
console.log("!")
}
It works because of custom valueOf method that is called when Object compared with primitive (such as Number). Main trick is that a.valueOf returns new value every time because it's calling exec on regular expression with g flag, which causing updating lastIndex of that regular expression every time match is found. So first time this.r.lastIndex == 0, it matches 1 and updates lastIndex: this.r.lastIndex == 1, so next time regex will match 2 and so on.

This is possible in case of variable a being accessed by, say 2 web workers through a SharedArrayBuffer as well as some main script. The possibility is low, but it is possible that when the code is compiled to machine code, the web workers update the variable a just in time so the conditions a==1, a==2 and a==3 are satisfied.
This can be an example of race condition in multi-threaded environment provided by web workers and SharedArrayBuffer in JavaScript.
Here is the basic implementation of above:
main.js
// Main Thread
const worker = new Worker('worker.js')
const modifiers = [new Worker('modifier.js'), new Worker('modifier.js')] // Let's use 2 workers
const sab = new SharedArrayBuffer(1)
modifiers.forEach(m => m.postMessage(sab))
worker.postMessage(sab)
worker.js
let array
Object.defineProperty(self, 'a', {
get() {
return array[0]
}
});
addEventListener('message', ({data}) => {
array = new Uint8Array(data)
let count = 0
do {
var res = a == 1 && a == 2 && a == 3
++count
} while(res == false) // just for clarity. !res is fine
console.log(`It happened after ${count} iterations`)
console.log('You should\'ve never seen this')
})
modifier.js
addEventListener('message' , ({data}) => {
setInterval( () => {
new Uint8Array(data)[0] = Math.floor(Math.random()*3) + 1
})
})
On my MacBook Air, it happens after around 10 billion iterations on the first attempt:
Second attempt:
As I said, the chances will be low, but given enough time, it'll hit the condition.
Tip: If it takes too long on your system. Try only a == 1 && a == 2 and change Math.random()*3 to Math.random()*2. Adding more and more to list drops the chance of hitting.

It can be accomplished using the following in the global scope. For nodejs use global instead of window in the code below.
var val = 0;
Object.defineProperty(window, 'a', {
get: function() {
return ++val;
}
});
if (a == 1 && a == 2 && a == 3) {
console.log('yay');
}
This answer abuses the implicit variables provided by the global scope in the execution context by defining a getter to retrieve the variable.

This is also possible using a series of self-overwriting getters:
(This is similar to jontro's solution, but doesn't require a counter variable.)
(() => {
"use strict";
Object.defineProperty(this, "a", {
"get": () => {
Object.defineProperty(this, "a", {
"get": () => {
Object.defineProperty(this, "a", {
"get": () => {
return 3;
}
});
return 2;
},
configurable: true
});
return 1;
},
configurable: true
});
if (a == 1 && a == 2 && a == 3) {
document.body.append("Yes, it’s possible.");
}
})();

Alternatively, you could use a class for it and an instance for the check.
function A() {
var value = 0;
this.valueOf = function () { return ++value; };
}
var a = new A;
if (a == 1 && a == 2 && a == 3) {
console.log('bingo!');
}
EDIT
Using ES6 classes it would look like this
class A {
constructor() {
this.value = 0;
this.valueOf();
}
valueOf() {
return this.value++;
};
}
let a = new A;
if (a == 1 && a == 2 && a == 3) {
console.log('bingo!');
}

I don't see this answer already posted, so I'll throw this one into the mix too. This is similar to Jeff's answer with the half-width Hangul space.
var a = 1;
var ａ = 2;
var а = 3;
if(a == 1 && ａ == 2 && а == 3) {
console.log("Why hello there!")
}
You might notice a slight discrepancy with the second one, but the first and third are identical to the naked eye. All 3 are distinct characters:
a - Latin lower case A
ａ - Full Width Latin lower case A
а - Cyrillic lower case A
The generic term for this is "homoglyphs": different unicode characters that look the same. Typically hard to get three that are utterly indistinguishable, but in some cases you can get lucky. A, Α, А, and Ꭺ would work better (Latin-A, Greek Alpha, Cyrillic-A, and Cherokee-A respectively; unfortunately the Greek and Cherokee lower-case letters are too different from the Latin a: α,ꭺ, and so doesn't help with the above snippet).
There's an entire class of Homoglyph Attacks out there, most commonly in fake domain names (eg. wikipediа.org (Cyrillic) vs wikipedia.org (Latin)), but it can show up in code as well; typically referred to as being underhanded (as mentioned in a comment, [underhanded] questions are now off-topic on PPCG, but used to be a type of challenge where these sorts of things would show up). I used this website to find the homoglyphs used for this answer.

Yes, it is possible! 😎
» JavaScript
if‌=()=>!0;
var a = 9;
if‌(a==1 && a== 2 && a==3)
{
document.write("<h1>Yes, it is possible!😎</h1>")
}
The above code is a short version (thanks to #Forivin for its note in comments) and the following code is original:
var a = 9;
if‌(a==1 && a== 2 && a==3)
{
//console.log("Yes, it is possible!😎")
document.write("<h1>Yes, it is possible!😎</h1>")
}
//--------------------------------------------
function if‌(){return true;}
If you just see top side of my code and run it you say WOW, how?
So I think it is enough to say Yes, it is possible to someone that said to
you: Nothing is impossible
Trick: I used a hidden character after if to make a function that its name is similar to if. In JavaScript we can not override keywords so I forced to use this way. It is a fake if, but it works for you in this case!
» C#
Also I wrote a C# version (with increase property value technic):
static int _a;
public static int a => ++_a;
public static void Main()
{
if(a==1 && a==2 && a==3)
{
Console.WriteLine("Yes, it is possible!😎");
}
}
Live Demo

JavaScript
a == a +1
In JavaScript, there are no integers but only Numbers, which are implemented as double precision floating point numbers.
It means that if a Number a is large enough, it can be considered equal to four consecutive integers:
a = 100000000000000000
if (a == a+1 && a == a+2 && a == a+3){
console.log("Precision loss!");
}
True, it's not exactly what the interviewer asked (it doesn't work with a=0), but it doesn't involve any trick with hidden functions or operator overloading.
Other languages
For reference, there are a==1 && a==2 && a==3 solutions in Ruby and Python. With a slight modification, it's also possible in Java.
Ruby
With a custom ==:
class A
def ==(o)
true
end
end
a = A.new
if a == 1 && a == 2 && a == 3
puts "Don't do this!"
end
Or an increasing a:
def a
#a ||= 0
#a += 1
end
if a == 1 && a == 2 && a == 3
puts "Don't do this!"
end
Python
You can either define == for a new class:
class A:
def __eq__(self, who_cares):
return True
a = A()
if a == 1 and a == 2 and a == 3:
print("Don't do that!")
or, if you're feeling adventurous, redefine the values of integers:
import ctypes
def deref(addr, typ):
return ctypes.cast(addr, ctypes.POINTER(typ))
deref(id(2), ctypes.c_int)[6] = 1
deref(id(3), ctypes.c_int)[6] = 1
deref(id(4), ctypes.c_int)[6] = 1
print(1 == 2 == 3 == 4)
# True
It might segfault, depending on your system/interpreter.
The python console crashes with the above code, because 2 or 3 are probably used in the background. It works fine if you use less-common integers:
>>> import ctypes
>>>
>>> def deref(addr, typ):
... return ctypes.cast(addr, ctypes.POINTER(typ))
...
>>> deref(id(12), ctypes.c_int)[6] = 11
>>> deref(id(13), ctypes.c_int)[6] = 11
>>> deref(id(14), ctypes.c_int)[6] = 11
>>>
>>> print(11 == 12 == 13 == 14)
True
Java
It's possible to modify Java Integer cache:
package stackoverflow;
import java.lang.reflect.Field;
public class IntegerMess
{
public static void main(String[] args) throws Exception {
Field valueField = Integer.class.getDeclaredField("value");
valueField.setAccessible(true);
valueField.setInt(1, valueField.getInt(42));
valueField.setInt(2, valueField.getInt(42));
valueField.setInt(3, valueField.getInt(42));
valueField.setAccessible(false);
Integer a = 42;
if (a.equals(1) && a.equals(2) && a.equals(3)) {
System.out.println("Bad idea.");
}
}
}

This is an inverted version of #Jeff's answer* where a hidden character (U+115F, U+1160 or U+3164) is used to create variables that look like 1, 2 and 3.
var a = 1;
var ﾠ1 = a;
var ﾠ2 = a;
var ﾠ3 = a;
console.log( a ==ﾠ1 && a ==ﾠ2 && a ==ﾠ3 );
* That answer can be simplified by using zero width non-joiner (U+200C) and zero width joiner (U+200D). Both of these characters are allowed inside identifiers but not at the beginning:
var a = 1;
var a‌ = 2;
var a‍ = 3;
console.log(a == 1 && a‌ == 2 && a‍ == 3);
/****
var a = 1;
var a\u200c = 2;
var a\u200d = 3;
console.log(a == 1 && a\u200c == 2 && a\u200d == 3);
****/
Other tricks are possible using the same idea e.g. by using Unicode variation selectors to create variables that look exactly alike (a︀ = 1; a︁ = 2; a︀ == 1 && a︁ == 2; // true).

Rule number one of interviews; never say impossible.
No need for hidden character trickery.
window.__defineGetter__( 'a', function(){
if( typeof i !== 'number' ){
// define i in the global namespace so that it's not lost after this function runs
i = 0;
}
return ++i;
});
if( a == 1 && a == 2 && a == 3 ){
console.log( 'Oh dear, what have we done?' );
}

Honestly though, whether there is a way for it to evaluate to true or not (and as others have shown, there are multiple ways), the answer I'd be looking for, speaking as someone who has conducted hundreds of interviews, would be something along the lines of:
"Well, maybe yes under some weird set of circumstances that aren't immediately obvious to me... but if I encountered this in real code then I would use common debugging techniques to figure out how and why it was doing what it was doing and then immediately refactor the code to avoid that situation... but more importantly: I would absolutely NEVER write that code in the first place because that is the very definition of convoluted code, and I strive to never write convoluted code".
I guess some interviewers would take offense to having what is obviously meant to be a very tricky question called out, but I don't mind developers who have an opinion, especially when they can back it up with reasoned thought and can dovetail my question into a meaningful statement about themselves.

If you ever get such an interview question (or notice some equally unexpected behavior in your code) think about what kind of things could possibly cause a behavior that looks impossible at first glance:
Encoding: In this case the variable you are looking at is not the one you think it is. This can happen if you intentionally mess around with Unicode using homoglyphs or space characters to make the name of a variable look like another one, but encoding issues can also be introduced accidentally, e.g. when copying & pasting code from the Web that contains unexpected Unicode code points (e.g. because a content management system did some "auto-formatting" such as replacing fl with Unicode 'LATIN SMALL LIGATURE FL' (U+FB02)).
Race conditions: A race-condition might occur, i.e. a situation where code is not executing in the sequence expected by the developer. Race conditions often happen in multi-threaded code, but multiple threads are not a requirement for race conditions to be possible – asynchronicity is sufficient (and don't get confused, async does not mean multiple threads are used under the hood).
Note that therefore JavaScript is also not free from race conditions just because it is single-threaded. See here for a simple single-threaded – but async – example. In the context of an single statement the race condition however would be rather hard to hit in JavaScript.
JavaScript with web workers is a bit different, as you can have multiple threads. #mehulmpt has shown us a great proof-of-concept using web workers.
Side-effects: A side-effect of the equality comparison operation (which doesn't have to be as obvious as in the examples here, often side-effects are very subtle).
These kind of issues can appear in many programming languages, not only JavaScript, so we aren't seeing one of the classical JavaScript WTFs here1.
Of course, the interview question and the samples here all look very contrived. But they are a good reminder that:
Side-effects can get really nasty and that a well-designed program should be free from unwanted side-effects.
Multi-threading and mutable state can be problematic.
Not doing character encoding and string processing right can lead to nasty bugs.
1 For example, you can find an example in a totally different programming language (C#) exhibiting a side-effect (an obvious one) here.

Here's another variation, using an array to pop off whatever values you want.
const a = {
n: [3,2,1],
toString: function () {
return a.n.pop();
}
}
if(a == 1 && a == 2 && a == 3) {
console.log('Yes');
}

Okay, another hack with generators:
const value = function* () {
let i = 0;
while(true) yield ++i;
}();
Object.defineProperty(this, 'a', {
get() {
return value.next().value;
}
});
if (a === 1 && a === 2 && a === 3) {
console.log('yo!');
}

Using Proxies:
var a = new Proxy({ i: 0 }, {
get: (target, name) => name === Symbol.toPrimitive ? () => ++target.i : target[name],
});
console.log(a == 1 && a == 2 && a == 3);
Proxies basically pretend to be a target object (the first parameter), but intercept operations on the target object (in this case the "get property" operation) so that there is an opportunity to do something other than the default object behavior. In this case the "get property" action is called on a when == coerces its type in order to compare it to each number. This happens:
We create a target object, { i: 0 }, where the i property is our counter
We create a Proxy for the target object and assign it to a
For each a == comparison, a's type is coerced to a primitive value
This type coercion results in calling a[Symbol.toPrimitive]() internally
The Proxy intercepts getting the a[Symbol.toPrimitive] function using the "get handler"
The Proxy's "get handler" checks that the property being gotten is Symbol.toPrimitive, in which case it increments and then returns the counter from the target object: ++target.i. If a different property is being retrieved, we just fall back to returning the default property value, target[name]
So:
var a = ...; // a.valueOf == target.i == 0
a == 1 && // a == ++target.i == 1
a == 2 && // a == ++target.i == 2
a == 3 // a == ++target.i == 3
As with most of the other answers, this only works with a loose equality check (==), because strict equality checks (===) do not do type coercion that the Proxy can intercept.

Actually the answer to the first part of the question is "Yes" in every programming language. For example, this is in the case of C/C++:
#define a (b++)
int b = 1;
if (a ==1 && a== 2 && a==3) {
std::cout << "Yes, it's possible!" << std::endl;
} else {
std::cout << "it's impossible!" << std::endl;
}

Same, but different, but still same (can be "tested" multiple times):
const a = { valueOf: () => this.n = (this.n || 0) % 3 + 1}
if(a == 1 && a == 2 && a == 3) {
console.log('Hello World!');
}
if(a == 1 && a == 2 && a == 3) {
console.log('Hello World!');
}
My idea started from how Number object type equation works.

An ECMAScript 6 answer that makes use of Symbols:
const a = {value: 1};
a[Symbol.toPrimitive] = function() { return this.value++ };
console.log((a == 1 && a == 2 && a == 3));
Due to == usage, JavaScript is supposed to coerce a into something close to the second operand (1, 2, 3 in this case). But before JavaScript tries to figure coercing on its own, it tries to call Symbol.toPrimitive. If you provide Symbol.toPrimitive JavaScript would use the value your function returns. If not, JavaScript would call valueOf.

I think this is the minimal code to implement it:
i=0,a={valueOf:()=>++i}
if (a == 1 && a == 2 && a == 3) {
console.log('Mind === Blown');
}
Creating a dummy object with a custom valueOf that increments a global variable i on each call. 23 characters!

This one uses the defineProperty with a nice side-effect causing global variable!
var _a = 1
Object.defineProperty(this, "a", {
"get": () => {
return _a++;
},
configurable: true
});
console.log(a)
console.log(a)
console.log(a)

By overriding valueOf in a class declaration, it can be done:
class Thing {
constructor() {
this.value = 1;
}
valueOf() {
return this.value++;
}
}
const a = new Thing();
if(a == 1 && a == 2 && a == 3) {
console.log(a);
}
What happens is that valueOf is called in each comparison operator. On the first one, a will equal 1, on the second, a will equal 2, and so on and so forth, because each time valueOf is called, the value of a is incremented.
Therefore the console.log will fire and output (in my terminal anyways) Thing: { value: 4}, indicating the conditional was true.

As we already know that the secret of loose equality operator (==) will try to convert both values to a common type. As a result, some functions will be invoked.
ToPrimitive(A) attempts to convert its object argument to a primitive
value, by invoking varying sequences of A.toString and A.valueOf
methods on A.
So as other answers using Symbol.toPrimitive, .toString, .valueOf from integer. I would suggest the solution using an array with Array.pop like this.
let a = { array: [3, 2, 1], toString: () => a.array.pop() };
if(a == 1 && a == 2 && a == 3) {
console.log('Hello World!');
}
In this way, we can work with text like this
let a = { array: ["World", "Hello"], toString: () => a.array.pop() };
if(a == "Hello" && a == "World") {
console.log('Hello World!');
}

Surprisingly, yes. The == loose equality operator in JS calls the valueOf() method of the object that's being compared. Therefore, you can create a class that returns an internal value, then increments that interval value every time it's called. Like this:
class AClass {
constructor(initalVal) {
this.val = initalVal;
}
valueOf() {
return this.val++;
}
}
const a = new AClass(1);
console.log(a==1 && a==2 && a==3)
I know that there are a lot of other answers to this question, but this is how you'd do it with ES6 syntax.
Note: If you don't want this to happen, then you should use the === operator to check for strict instead. Like this:
class AClass {
constructor(initalVal) {
this.val = initalVal;
}
valueOf() {
return this.val++;
}
}
const a = new AClass(1);
console.log(a===1 && a===2 && a===3)

Yes, you can Do that, see the following JavaScript code:
let a = 0 // Create a variable and give it a value
if( a !== 1 && a !== 2 && a !== 3 )
{
console.log("true")
}
Explanation of the solution:
Simply , we add the not equal sign
before the == sign so that we tell the language that these values are
not equal to the value in the variable

Variable reassignment with chained booleans or ternary operator [closed]

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 2 years ago.
Improve this question
To clarify:
This is purely for experimental purposes, to learn the quirks, odds and ends of a new (to me) language. I would of course write it readable if I ever were to share this code with anyone else. :-)
I have a function someFunction(x), and two global variables:
let m = 2;
let e = 5;
Inside the function, I want to check if x == m. If this is true, I'd like to pass m to a side function call (sideFunction(m)), then reassign x to e to complete someFunction.
This does work as expected:
const someFunction = x => {
if (x == m) {
sideFunction(m);
x = e;
}
doOtherStuffWith(x);
}
However, I'd like to shorten it, preferably to one line. This is also to understand more about ternaries and/or boolean chaining.
I have tried these two methods:
// Boolean chaining
const someFunction = x => {
x == m && sideFunction(m) && (function () {x = e})();
doOtherStuffWith(x);
}
This does not work, presumably because the assignment x = e only applies to the x in the local scope of the inner, anonymous function...?
// Ternary operator
const someFunction = x => {
x = (x == m && sideFunction(m)) ? e : x;
doOtherStuffWith(x);
}
This does not work, presumably because sideFunction(m) doesn't actually get called, for some reason...?
How can I fix these to make them work?
Alternatively, are there other, elegant ways to perform this check/call/reassignment without a full multi-line if block?
Thank you very much!

The problem with
x == m && sideFunction(m) && (function () {x = e})();
is that && evaluates left-to-right, and will stop as soon as the first falsey value is found. Unless sideFunction returns something explicitly truthy, the third IIFE:
(function () {x = e})()
will never run, resulting in x never being reassigned.
x is local in that function. If you can get the function to run, it will reassign x as desired.
You could use the comma operator:
x == m && (sideFunction(m), x = e);
Similarly
x = (x == m && sideFunction(m)) ? e : x;
won't work because sideFunction would have to return something truthy for the left side of the conditional to evaluate truthily - otherwise, x will be assigned to x, no change.
All this said - I'd highly recommend not doing any of these. Your first approach is much more readable, and readability is much more important than line conservation.

You could take a compound expression with a comma operator and do not care about the result value of sideFunction.
const fn = x => (x == m && (sideFunction(m), x = e), doOtherStuffWith(x));

The comma operator , can be used to chain expressions and return the last one e.g.,
var x = (1 + 2, 10 + 2, 100 + 2);
x;
//=> 102
We can use it to evaluate sideFunction(m) and return e: (sideFunction(m), e).
Since you always want to execute doOtherStuffWith you only need to work out whether to give it e or the original x:
const someFunction = x => doOtherStuffWith(x == m ? (sideFunction(m), e) : x);

Javascript If Else always picks option 1

So my Javascript code is supposed to randomly pick, but it always picks the first option, even if I switch the options, it always goes to the one on top. Did I make an error, or how can I get this to work?
function test() {
var values = ["1","2"],
valueToUse = values[Math.floor(Math.random() * values.length)];
// do something with the selected value
alert(valueToUse);
}
if (values = 1) {
window.location = "tail.html"; }
else if (values = 2) { window.location = "head.html"; }

if (values = 1) {
That's an assignment, not a comparison (===). It will always be true.
Additionally, you never call test().
Additionally, you should be testing valueToUse not values, since that is where you assign your random choice.
Additionally, both valueToUse and values are locally scoped to the test() function and are not available outside it. Your assignment creates a new, global variable.

You need to use == (or ===) instead of =.
if (values == 1) {
// and so on...
}

To make your code fly, you could return a random selected value from the array, store it in value and check it for the new direction. I omit window.location and us simply console.log for displaying the result of the comparison.
The comparison is strict, with a check for a string.
function test() {
var values = ["1", "2"],
valueToUse = values[Math.floor(Math.random() * values.length)];
return valueToUse;
}
var value = test();
if (value === '1') {
console.log( "tail.html");
} else if (value === '2') {
console.log( "head.html");
}

Well, for the start if (values = 1) is not the valid way to compare.
= annotates assignment so basically you are setting the value to 1. If you want to compare values in js, you do it this way if (values === 1).
In other programming language u usually use == instead however in js if you compare it this way it will only check the value not the type of the variable so a number 2 will match a string "2"
if (2 == "2") {//result is true}.
If you use === it will check both the value and the type
if (2 == "2") {//result is false}

setting a javascript variable with an if statement -- should the 'var = x' be inside or outside the IF?

This may be a basic question but I'm having difficulty finding an answer.
You want to set var B based on var A
would you do
var B = if(A == "red"){"hot"}else{"cool"}
I don't think this works.
I guess you could do
if(A == "red"){var B = "hot"}else{var B = "cool"}
This doesn't seem particularly elegant. I mean I would prefer something that starts with var b = .... just for clarity's sake.

perfect use for a ternary
var B = (A ==="red") ? "hot":"cool";
Ternary expressions will always return the first value if true, the second value if not. Great for one-off if/else statements, but if you get into more nested conditions, be sure to use the traditional if/else blocks for readability.

While you could use an if statement for this, you would probably be best off using the ternary operator. In the example you describe, you'd set it up as:
var B = A === "red" ? "hot" : "cool";
The ternary operator in general works as follows:
someBooleanExpression ? resultIfSomeBooleanExpressionIsTrue : resultIfSomeBooleanExpressionIsFalse;
Also note the use of === to check comparison, rather than a single =, which is assignment. == could also be used, but might end up being less accurate

Initialize your variables first, then compare using if statement and assign new values where necessary
var a = "red";
var b;
if(a=="red"){
b="hot";
}
else{
b="cold";
}

In your core you are declaring an initializing a to "red" which is truthy so the else clause will never be used. In JavaScript it's an if-statement not an if-expression as in say F#. That means you are not initializing B.
What you are probably looking for is
var B = A ==="red" ? "hot" : "cool"
That is to say use the conditional operator (*) to initialize B to "hot" if A is a string that equals "red" otherwise to "cool"
(*) The operator is sometimes imprecisely called the ternary operator or more mistakenly the tertiary. It isn't so bad to call it the ternary operator as long as there's never added another ternary operator, at which point it would be ambiguous

First option
var A = "blue";
var B = (A == "red") ? "hot" : "cool";
Second option
var A = "blue";
var B = ""; // Or simply: var B;
if (A == "red") B = "hot"; else B = "cool";