Why Does ParseInt("014") Come Out To 12?

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

Solution 1:

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).

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• See for more information

Solution 2:

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);

Solution 3:

Leading zeros make the number octal

Solution 4:

It's an octal number

8 + 4 == 12