Javascript Array.find method sometimes returns item, sometimes undefined

Question

I have an array that looks like this:

const list = [["GET", /github/g]];

It's an array with a sub array, index 0 having a string, index 1 having a regular expression.

I'm trying to do something like this:

function checkList() {
    return list.find(i => i[0] === "GET" && i[1].test("github.com"));
}

Which I would expect to return the list it finds every time, but that's not the case, it flips back and forth from returning the list and returning undefined.

Here's an example:

let i = 1;
const list = [["GET", /github/g]];
while (i <=10) {
    console.log(list.find(i => i[0] === "GET" && i[1].test("github.com")))
    i++
}

In this example I just loop through it and log the results to show that it returns the list and undefined every other time. What's going on here?

user56828cc2 · Accepted Answer · 2020-05-18 08:43:39Z

1

This is happening because the RegExp in i[1] is actually stateful. Your code calls i[1].test(...) on odd iterations, matching against "github" in "github.com", and advancing lastIndex. Then, on the next even iteration, the next call to test fails to find another match, reaching the end of the string. After that, the RegExp restarts the search from the beginning of the string.

If you would just like to check whether the pattern is found at least once, you could simply remove the /g flag at the end.

From MDN docs:

As with exec() (or in combination with it), test() called multiple times on the same global regular expression instance will advance past the previous match.

(See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/test)

edited May 18, 2020 at 8:43

answered May 18, 2020 at 8:41

user56828cc2

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4 Comments

Quyet Nguyen Over a year ago

As an another solution, he might need to use string.match(regex) instead.

user56828cc2 Over a year ago

True, although, I don't immediately see a need for a global match in this case anyway. I'd lean more towards not using features that aren't necessary.

Joshua Terrill Over a year ago

I'm building a library that other people are going to use, so I can't just remove the /g because that might be something that someone wants to use in a url pattern. However, @QuyetNguyen 's solution of changing it to be match appeared to do the trick. Thank you.

user56828cc2 Over a year ago

@QuyetNguyen, you should probably post it as an answer

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