Javascript: what's the point of RegExp.compile()?

I've got a situation where I want to get a regexp from the user and run it against a few thousand input strings. In the manual I found that the RegExp object has a .compile() method which is used to speed things up ins such cases. But why do I have to pass the regexp string to it again if I already passed them in the constructor? Perhaps constructor does the compile() itself?

Answers:

Answer

The RegExp().compile() method is deprecated. It's basically the same as the constructor, which I assume is why it was deprecated. You should only have to use the constructor nowadays.

In other words, you used to be able to do this:

var regexp = new RegExp("pattern");
regexp.compile("new pattern");

But nowadays it is not any different from simply calling:

var regexp = new RegExp("pattern");
regexp = new RegExp("new pattern");
Answer

And with Opera 11, running RegExp.compile() will actually cause errors.

Evidently, when Opera "compiles" a regex, it wraps the re.source string in forward slashes (e.g. re.source == "^(.)" becomes "/^(.)/"). If you manually compile the regex, Opera doesn't recognize this fact and goes ahead and compiles it again (re.source becomes "//^(.)//"). Each compile results in an extra set of forward slashes, which changes the meaning of the regular expression and results in errors.

Answer

You have to compile your regex first to use it if you are using /, try this out:

var regex=new RegExp('/[a-zA-Z]/')

console.log("not compiled with escape /", regex.test("ciao") )

regex.compile()

console.log("compiled", regex.test("ciao") )

var regex=new RegExp('[a-zA-Z]')

console.log("not compiled, but no escape /", regex.test("ciao") )

Answer

As far as i can tell all RegExp.compile does is replace the underlying regular expression of a RegExp object. I think compile may have had value in the past, but all modern JS engines "compile" the regex on first call and cache that "compiled" version.

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