Possible Duplicate:
Workarounds for JavaScript parseInt octal bug
It seems as though leading zeroes should just be ignored when parsing for an Int. What is the rationale behind this?
Asked
Active
Viewed 590 times
3
4 Answers
16
It is parsed as octal number, you need to specify base too:
parseInt("014", 10) // 14
Quoting:
If the input string begins with "0x" or "0X", radix is 16 (hexadecimal).
If the input string begins with "0", radix is eight (octal). This feature is non-standard, and some implementations deliberately do not support it (instead using the radix 10). For this reason always specify a radix when using parseInt.
If the input string begins with any other value, the radix is 10 (decimal).
Sarfraz
- 377,238
- 77
- 533
- 578
11
Because it is parsed as an octal number, and not decimal. From MDC:
- If the input string begins with "0x" or "0X", radix is 16 (hexadecimal).
- If the input string begins with "0", radix is eight (octal). This feature is non-standard, and some implementations deliberately do not support it (instead using the radix 10). For this reason always specify a radix when using parseInt.
- If the input string begins with any other value, the radix is 10 (decimal).
To force it to parse as Decimal, just supply 10 as your second argument (base).
var i = parseInt(012,10);
TJHeuvel
- 12,403
- 4
- 37
- 46
-
Is there a way to force it to parse as decimal? – T Nguyen Sep 08 '11 at 11:28
-
To solve it you need to a radix of 10 read more on: https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/parseInt – voigtan Sep 08 '11 at 11:29
-
See my edit, or @Sarfraz's answer – TJHeuvel Sep 08 '11 at 11:35
