should I drop the 'new' in this function passing regexp?
This question already has an answer here:
RegExp("foo") and new RegExp("foo") do the same thing:
15.10.3 The RegExp Constructor Called as a Function
15.10.3.1 RegExp(pattern, flags)
If pattern is an object R whose [[Class]] internal property is "RegExp" and flags is undefined, then return R unchanged. Otherwise call the standard built-in RegExp constructor (15.10.4.1) as if by the expression new RegExp( pattern, flags) and return the object constructed by that constructor.
http://es5.github.io/#x15.10.3
In simpler terms, RegExp(/someRegex/) returns that regex, while RegExp("someString") creates a new regexp from the string, just as new RegExp does.
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