Moderator note: Please resist the urge to edit the code or remove this notice. The pattern of whitespace may be part of the question and therefore should not be tampered with unnecessarily. If you are in the "whitespace is insignificant" camp, you should be able to accept the code as is.
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.
==
when you mean ===
, have a coding standard that bans non-ASCII variable names, and have a linting process which enforces the previous two morals. - anyone 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.
Answered 2023-09-20 20:26:35
valueOf()
operation? - anyone valueOf
is slightly better. - anyone i
doesn't bother the engine. ;) - anyone 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!")
}
Answered 2023-09-20 20:26:35
var ᅠ2 = 3
has been used; so there are the three variables aᅠᅠ= 1, ᅠ2 = 3, a = 3
(a␣ = 1, ␣2 = 3, a = 3
, so that (a␣==1 && a==␣2 && a==3)
)… - anyone 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.");
}
Answered 2023-09-20 20:26:35
with
." - anyone with
is not allowed. - anyone with
so it can happen - anyone ==
. And ===
prevents the accepted answer - anyone ==
but I haven't seen with
since ... well actually never outside of JS documentation where it says "please don't use that". Anyway, a nice solution. - anyone 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);
Answered 2023-09-20 20:26:35
without valueOf
, well... its more indirect but basically the same thing. - anyone toString
or valueOf
but this one caught me completely out of guard. Very clever and I didn't know it did call .join
internally, but it makes total sense. - anyone 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;
}
}
}
Answered 2023-09-20 20:26:35
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.
Answered 2023-09-20 20:26:35
exec
again will begin searching from that index. MDN is not very clear. - anyone this.r
regex object remember the state / index. Thanks! - anyone exec
though, not an integer to be stringified. - anyone 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.
Answered 2023-09-20 20:26:35
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.
Answered 2023-09-20 20:26:35
a
is a property of this
which it does not appear to be. If a
was a local variable (which it looks like), then this would not work. - anyone a == 1
implies than a
is a variable somewhere, not a property of this
. While there is an oddball place like globals where both could be true, generally, declaring a variable with var a
or let a
means there's no this
that lets you access a
as a property like you're code assumes. So, your code is apparently assuming some weird global variable thing. For example, your code does not work in node.js and not in strict mode inside a function. You should specify the exact circumstances where it works and probably explain why it works. Otherwise, it's misleading. - anyone a
is not a local variable and is defined on the global scope with an incrementing getter. - anyone 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.");
}
})();
Answered 2023-09-20 20:26:35
===
, not just ==
. - anyone this
being the global object inside the body of the arrow function. - anyone (a == 3 && a == 2 && a == 1)
? - anyone 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!');
}
Answered 2023-09-20 20:26:35
function A() {value = 0;
at the start? - anyone valueOf
is being overridden, this method is usually called automatically by JavaScript behind the scenes, and not explicitly in code
so when we compare the value it actually increments a.. - anyone 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.
Answered 2023-09-20 20:26:35
a
: a︀
a︁
a︂
. No more worrying about discrepancies. - anyone 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 toif
. In JavaScript we can not override keywords so I forced to use this way. It is a fakeif
, but it works for you in this case!
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!😎");
}
}
Answered 2023-09-20 20:26:35
if=()=>!0
- anyone document.write
? That's a surefire way not to get hired regardless of the rest of the answer. - anyone console.log
but I changed it to document.write. Really always I use console.log
in my codes but here I just want show a text to users in StackOverflow code snippet box. So I wanted to show my message more beautiful than the message generated by console.log
. Click the Run Code Snippet
button on my answer and on other answers. The SO Code Snippet let me to use html and JS and CSS then I wanted to use it in my answer and make it nice. I think It has not any negative side effect and did not made my answer large or complected. - anyone In JavaScript, there are no integers but only Number
s, 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.
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.
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
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
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.");
}
}
}
Answered 2023-09-20 20:26:35
Integer a = 42
(or does it)? As I understand autoboxing, Integer a = 42; a == 1 && a == 2 && a == 3
should box all of the ints. Or does this unbox a for the comparisons? - anyone Integer == int
seems to result in unboxing. But using Integer#equals(int)
forces autoboxing, so it works. Thanks for the comment! - anyone Numbers
in JS, which are basically like double
s. They can look like integers and you can use them like integers, but they still aren't integers. I don't think n == n + 1
can ever be true for integers in Java/Python/C/Ruby/... - anyone 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
).
Answered 2023-09-20 20:26:35
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?' );
}
Answered 2023-09-20 20:26:35
__defineGetter__
is actually not part of the js language, just an ugly version of defineProperty
. typeof
is not a function and this undeclared i
is just awful. Still seems to be worth 40 upvotes :/ - anyone __defineGetter__
is deprecated per developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… but it clearly executes in my FireFox v 57.0.4 so I opted to show this instead of defineProperty()
because legacy code is real and cannot be ignored. Regardless of the ugliness, declaring i
in the way I did is a well known/documented behavior. Maybe I was just in a PCG mood ¯\_(ツ)_/¯ - anyone 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.
Answered 2023-09-20 20:26:35
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:
1 For example, you can find an example in a totally different programming language (C#) exhibiting a side-effect (an obvious one) here.
Answered 2023-09-20 20:26:35
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');
}
Answered 2023-09-20 20:26:35
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!');
}
Answered 2023-09-20 20:26:35
this
being the window object) - anyone 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:
{ i: 0 }
, where the i
property is our countera
a ==
comparison, a
's type is coerced to a primitive valuea[Symbol.toPrimitive]()
internallya[Symbol.toPrimitive]
function using the "get handler"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.
Answered 2023-09-20 20:26:35
Symbol.toPrimitive
in the same way on an object would work just as well. - anyone 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;
}
Answered 2023-09-20 20:26:35
&&
for logical "and". - anyone 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.
Answered 2023-09-20 20:26:35
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
.
Answered 2023-09-20 20:26:35
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!
Answered 2023-09-20 20:26:35
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)
Answered 2023-09-20 20:26:35
a
: get: (a => () => ++a)(0),
no global necessary. - anyone 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.
Answered 2023-09-20 20:26:35
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 ofA.toString
andA.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!');
}
Answered 2023-09-20 20:26:35
If we use JavaScript's property of converting objects to primitive values and its getter functions, it can be possible.
const a = {
value: 0,
valueOf: function() {
return this.value += 1;
}
}
if (a == 1 && a == 2 && a == 3) {
console.log('it can')
}
Answered 2023-09-20 20:26:35
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
Answered 2023-09-20 20:26:35