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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?
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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
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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
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Is there a way to force it to parse as decimal? – T Nguyen Sep 08 '11 at 11:28
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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
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See my edit, or @Sarfraz's answer – TJHeuvel Sep 08 '11 at 11:35
