Why parseInt(0.0000005) returns 5 in JavaScript

Tari Ibaba

Last updated on July 02, 2024

Why parseInt(0.0000005) returns 5 in JavaScript

We can't believe our eyes when we see it:

You jump out of your skin and my mouth hangs open for over 10 minutes.

Just when we thought we knew everything about parseInt and the idiotic things only JavaScript did?

The funny thing is it returns 0 for four and five zeroes after the decimal point.

But just add one more zero and it starts acting crazy?

The difference from Number is even more stark.

✅ parseInt vs Number -- four and five zeroes:

But:

❌ parseInt vs Number -- six zeroes:

But we figure it out eventually.

parseInt: The missing parts

By understanding how the parseInt function work:

It takes two arguments: the string and the base or radix.

parseInt(string)

parseInt(string, radix)

Look what it does for 6 zeroes when it's a string:

parseInt('0.5') // 0
parseInt('0.05') // 0
parseInt('0.005') // 0
parseInt('0.0005') // 0
parseInt('0.00005') // 0
parseInt('0.000005') // 0
// 6
parseInt('0.0000005') // 0
parseInt('015') // 15
parseInt('015', 8) // 13

But for Number:

Number('0.5') // 0.5
Number('0.05') // 0.05
Number('0.005') // 0.005
Number('0.0005') // 0.0005
Number('0.00005') // 0.00005
Number('0.000005') // 0.000005
// 6 !
Number('0.0000005') // 0.0000005
Number('015') // 15

But what happens when it's a number?

A number like 0.0000005?

Step 1: Convert the number to a string

Number gets coerced to a string!

So look what parseInt does to 0.0000005:

String(0.5);      // => '0.5'
String(0.05);     // => '0.05'
String(0.005);    // => '0.005'
String(0.0005);   // => '0.0005' 
String(0.00005);  // => '0.00005'
String(0.000005); // => '0.000005'

String(0.0000005); // => '5e-7' 👈 this is important

Step 2: Do the actual rounding

So:

parseInt(0.0000005) → parseInt('5e-7')

And now see what happens for '5e-7' for parseInt:

Why?

Because of how parseInt works: It interprets only the leading portion of the string as an integer value.

It doesn't recognize e, so it ignores it and everything onwards.

This is same reason why:

parseInt(999999999999999999999) → parseInt(1e+21) → 1
parseInt(0.0000007) → parseInt(7) → 7

So Number is your best bet if you wanna avoid surprises when converting a string to a number

And if you need just the integer part, Math.floor() is here for you:

Math.floor(0.5);      // 0
Math.floor(0.05);     // 0
Math.floor(0.005);    // 0
Math.floor(0.0005);   // 0
Math.floor(0.00005);  // 0
Math.floor(0.000005); // 0
Math.floor(0.0000005); // 0 ✅ perfect

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