How do I work around JavaScript's parseInt octal behavior?

Question

Try executing the following in JavaScript:

parseInt(\'01\'); //equals 1
parseInt(\'02\'); //equals 2
parseInt(\'03\'); //equals 3
parseInt(\'04\'); //equals          


        

Answer 1

Would it be very naughty to replace parseInt with a version that assumes decimal if it has no second parameter? (note - not tested)

parseIntImpl = parseInt
parseInt = function(str, base){return parseIntImpl(str, base ? base : 10)}

Answer 2

Specify the base:

var number = parseInt(s, 10);

Answer 3

If you know your value will be in the signed 32 bit integer range, then ~~x will do the correct thing in all scenarios.

~~"08" === 8
~~"foobar" === 0
~~(1.99) === 1
~~(-1.99)  === -1

If you look up binary not (~), the spec requires a "ToInt32" conversion for the argument which does the obvious conversion to an Int32 and is specified to coerce NaN values to zero.

Yes, this is incredibly hackish but is so convenient...

Answer 4

If you've done a bunch of coding already with parseInt and don't want to add ",10" to everything, you can just override the function to make base 10 the default:

window._oldParseInt = window.parseInt;
window.parseInt = function(str, rad) {
    if (! rad) {
        return _oldParseInt(str, 10);
    }
    return _oldParseInt(str, rad);
};

That may confuse a later reader, so making a parseInt10() function might be more self-explanatory. Personally I prefer using a simple function than having to add ",10" all the time - just creates more opportunity for mistakes.