parseInt outputting int instead of octal number

Question

I've been following a course in Treehouse about JavaScript. I'm currently on the parseInt method. In one of their example, this code was shown:

var j = parseInt("012");

saving it and running in Chrome's console resulted to 10 in their example (which is understandable as 012 is 10 in octal), but in my own browser, it resulted to 12. Is it correct or am I doing something wrong?

Simple solution here. When using parseInt(), ALWAYS pass the desired radix as the second argument. Always. — jfriend00
– jfriend00, Commented May 12, 2015 at 0:30

jdphenix · Accepted Answer · 2015-05-12 00:17:20Z

4

It's not strange - the behavior is actually implementation dependent.

It could parse as octal 012 or decimal 12.

It's always best to specify the radix parameter when using parseInt(), for example parseInt('012', 10).

answered May 12, 2015 at 0:17

jdphenix

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1 Comment

defc0de Over a year ago

Thanks, I actually specify the radix whenever I use parseInt, I was just confused why the tutorial video had a different browser output when I ran the same code.

floribon · Accepted Answer · 2015-05-12 00:22:51Z

0

The difference comes from the type of your variable.

A Number 012 is considered octal, so parseInt(012) is 10.

A String "012" is nothing until parsed. By default parseInt will consider a string as a decimal value, hence "012" will be parsed as 12.

parseInt uses the decimal base by default, but you can force it to the octal base if you like:

parseInt("012", 8) === parseInt(12, 8) === 10
parseInt( 012 , 8) === parseInt(8, 10) === 8

answered May 12, 2015 at 0:22

floribon

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