Why does parseInt(013) return 11?

Question

Possible Duplicate:
Workarounds for JavaScript parseInt octal bug

Here's a jsfiddle that shows the behavior:

Pretty simple question, any ideas?

Code is just:

parseInt(013)

https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Values,_variables,_and_literals#Integers — deceze, Nov 28 '12 at 19:15

score 12 · Accepted Answer · answered Nov 28 '12 at 19:15

12

Because if your number is starting with '0', then it's considered as octal, thus

'013' = 1 * 8 + 3 = 11

answered Nov 28 '12 at 19:15

Orabîg

score 10 · Answer 2 · answered Nov 28 '12 at 19:14

10

parseInt() expects a string. You have provided an octal, 013.

Use:

parseInt('013', 10)

Note: I would also encourage passing radix, for clarity.

answered Nov 28 '12 at 19:14

Jason McCreary

score 2 · Answer 3 · answered Nov 28 '12 at 19:17

2

The number 013 is being interpreted as an octal. It has nothing to do with parseInt; in fact, var a = 013; will have a be 11.

answered Nov 28 '12 at 19:17

Ivan

3 Answers3