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

You may also, instead of using parseFloat or parseInt, use the unary operator (+).

+"01"
// => 1

+"02"
// => 2

+"03"
// => 3

+"04"
// => 4

+"05"
// => 5

+"06"
// => 6

+"07"
// => 7

+"08"
// => 8

+"09"
// => 9

and for good measure

+"09.09"
// => 9.09

MDN Link

The unary plus operator precedes its operand and evaluates to its operand but attempts to convert it into a number, if it isn't already. Although unary negation (-) also can convert non-numbers, unary plus is the fastest and preferred way of converting something into a number, because it does not perform any other operations on the number.