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Why Does JavaScript’s parseInt(0.0000005) Print “5”? 🤔

Jagroop Singh on November 10, 2024

Why Does JavaScript’s parseInt(0.0000005) Print “5”? 🤔 JavaScript’s parseInt() function is quite handy for converting strings into integ...

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Mike Talbot ⭐ • Nov 11

It's personally why I convert things into numbers using + rather than parsing. I don' have to remember or even know if it's a number or a string. My preference:

    const number = Math.floor(+someValueIWantToConvertToAnInt)

    // or often

   if(+someValue > 1) {}

Jagroop Singh • Nov 11

Nice trick @miketalbot ,
Using the + simplifies conversion by implicitly handling type coercion, making the code more concise without needing explicit parsing.

KMohZaid • Nov 13

Thankssaa

@whereisthebug • Nov 10

Nice catch!

As you correctly pointed out, parseInt converts the number to a string. And JavaScript converts numbers to strings using scientific notation when the magnitude of the number (i.e. the number without the sign) is larger or equal than 1000000000000000000000 (21 zeroes) or smaller or equal than 0.0000001 (7 zeroes.)

Jagroop Singh • Nov 10

Exactly! @whereisthebug Do you know any other tricky things about JS?

Shifa Ur Rehman • Nov 11

What do you mean? I thought the whole javascript was a tricky thing 😂

Jagroop Singh • Nov 11

Ben Sinclair • Nov 13

This is another one of those issues which isn't an issue if you use the function as it's defined, and it's defined to take a string parameter. There's no reason I can think of to pass a float into it, you'd use an explicit cast or a dedicated function like Math.floor for that purpose.

Jagroop Singh • Nov 13

yes exactly !!

john • Nov 10

I don't know what's happening in JS. It is javascript bug or something else, okay I understand the concept of parseInt(0.0000005) thanks for explaining that but why console.log(0.1 + 0.2 == 0.3) returns False.

Also I think developers must know these edge cases points of js because somehow it will used in case of payments then it really hard to debug for begineers to debug

Jagroop Singh • Nov 10

@john12 ,
The behavior you're noticing isn't a bug, but a well-known quirk in JavaScript (and many other programming languages) related to how it handles floating-point arithmetic.

Peter Vivo • Nov 11 • Edited

At least

parseInt("0.0000005")
// => 0

but that is great catch I don't know about it ( I was never use parseInt )

Jagroop Singh • Nov 11

Yes, parseInt("0.0000005") returns 0 because parseInt stops parsing at the decimal point.
So , I thinks it's good to convert numbers to string before parsing into integer.

Web • Nov 10

That's cool, I already know about second one , what about this

console.log(null == undefined);

Jagroop Singh • Nov 10

That one is easy @works ,
== means loose equality that convert both values to a common type before comparing. So it returns true

If we usr === then it returns false

Web • Nov 10

Yes correct.

Boniface Gordian • Nov 13

This is such an interesting topic—parseInt() is one of those quirks in JavaScript that always surprises people! I really liked how you explained the behavior with the radix and how it defaults to 10 in most cases. It’s definitely one of those ‘gotcha’ moments for developers. Personally, I try to stick with Number() for clarity, but there are definitely cases where parseInt() shines. Have you ever run into unexpected issues with it in real-world projects? Great work on the article!

Jagroop Singh • Nov 13

Thanks @boniface_gordian

A. Merezhanyi • Nov 12 • Edited

That would be great you’d started with the history and mentioned that parseInt was designed to parse stings to numbers for cases like “10px” -> “10”, that explains a lot.
That worth mentioning that according to IEEE 754 all other programming languages do the same math and add a link to 0.30…4.com

Mike Stemle • Nov 11

The 0.1+0.2===0.3 isn't a JavaScript issue, that's an IEEE 754 floating point math error. This is pretty common to a number of languages including JavaScript, Python, Perl, C/C++, etc.

Bernard Poulin • Nov 11

parseInt() is indeed handy when used correctly. Calling it with a number instead of a string would be considered a mistake.
If the intention was to convert a float into an integer, the popular quick hack is |0 as in 0.0000005|0. The more official way is to use Math.floor() obviously. parseInt() is simply the wrong method to do this.
Also note that negative values might or might not do what you'd expect. Depends on the use case.
I understand that the logic was maybe expecting a string and suddenly a number comes along at runtime and it was "working most of the time". The solution from Mike Talbot is indeed the best if you need to counter this issue. You can of course still combine it with the |0 hack

+0.0000005|0

peerreynders • Nov 15

Bitwise OR is a flawed solution as its result is always a signed 32 bit integer (true of all bitwise operators except unsigned right shift which produces an unsigned 32 bit integer).

JavaScript's safe range for integers is much larger; Number.MAX_SAFE_INTEGER to Number.MIN_SAFE_INTEGER:

let max = 2_147_483_647;
let min = ~max;
console.log(max | 0); //  2147483647 ✔
console.log(min | 0); // -2147483648 ✔

max += 1;
min -= 1;
console.log(max | 0); // -2147483648 ✘
console.log(min | 0); //  2147483647 ✘
console.log(Math.trunc(max)); //  2147483648 ✔
console.log(Math.trunc(min)); // -2147483649 ✔

max = Number.MAX_SAFE_INTEGER;
min = Number.MIN_SAFE_INTEGER;
console.log(max); //  9007199254740991 ✔
console.log(min); // -9007199254740991 ✔

max += 2;
min -= 2;
console.log(max); //  9007199254740992 ✘ (i.e. loss of precision)
console.log(min); // -9007199254740992 ✘

// Bonus: More on loss of precision
const update = (value: number): [number, number] => [value, Math.trunc(value)];
let data = update(5.000000000000001);

const show = ([left, right]: [number, number]) => [
  left,
  Number.isInteger(left),
  right,
  Number.isInteger(right),
];
console.log.apply(undefined, show(data));
// 5.000000000000001,  false,  5,  true

data = update(5.0000000000000001);
console.log.apply(undefined, show(data));
// 5,  true,  5,  true

And personally I prefer the explicitness of Number() coercion (e.g. Math.trunc(Number(0.0000005))) over the terseness of the unary plus operator.

Jagroop Singh • Nov 16

Thanks for sharing your views.

José Pablo Ramírez Vargas • Nov 11

It's your own fault as a developer. The function's name clearly states it will return an integer. You cannot possible expect a floating-point value to come out correctly from parse*Int*.

Nikunj Bhatt • Nov 11

I hope you are just joking.

José Pablo Ramírez Vargas • Nov 11

Why? Is there a punch line in what I wrote? Or is it normal to expect a floating-point value from a function named parseInt, especially when there's another function called parseFloat? What's confusing here?

Daniel Santana • Nov 11

I laughed at first as well because I was thinking with JS mindset. And for anyone with a bit of experience with JS that is indeed funny.
The parseInt function is supposed to receive anything and try to return an int from it.

So yes, most of the time the input is a non int value that will return an int, that’s why Jose’s first comment sound like a sarcastic joke.

But my guess is that the name of the function is not so specific and one could understand it as a function that receives and int then parses it into something else.

But what makes it funny is the arrogant tone of blaming who wrote the code based on a wrong understanding of what the function does. 😂

It’s funny if it is a joke, but even funnier because it’s not 😆

José Pablo Ramírez Vargas • Nov 11

Ah, I see. FYI and full disclosure: I don't blame the author directly when I said "you as a programmer". "You" here refer to anybody that stumbled into this problem expecting a float out of parseInt, hehe.

Vadim • Nov 17

Becase use parseFloat()