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identifier-identification

v0.1.1

Published

A JavaScript prollyfill for the proposed `String.isIdentifierStart` and `String.isIdentifierPart` methods.

Readme

JavaScript identifier identification Build status Dependency status

A JavaScript prollyfill for the proposed String.isIdentifierStart and String.isIdentifierPart methods, based on the March 8, 2013 draft of the strawman.

Feel free to fork if you see possible improvements!

Installation

In a browser:

<script src="identifier.js"></script>

Via npm:

npm install identifier-identification

Then, in Node.js:

require('identifier-identification');

Usage

// Is U+0B83 a valid `IdentifierStart` code point as per ECMAScript 3?
String.isIdentifierStart(0x0B83, 3);
// → false

// What about ECMASCript 5?
String.isIdentifierStart(0x0B83, 5);
// → false

// Ok, and in ECMAScript 6?
String.isIdentifierStart(0x0B83, 6);
// → true

// Is U+2FA1D a valid `IdentifierPart` code point as per ECMAScript 3?
String.isIdentifierPart(0x2FA1D, 3);
// → false

// What about ECMASCript 5?
String.isIdentifierPart(0x2FA1D, 5);
// → false

// Ok, and in ECMAScript 6?
String.isIdentifierPart(0x2FA1D, 6);
// → true

Notes

In order to make this polyfill ES3/ES5-compatible, this script includes a String.prototype.codePointAt polyfill.

If the second, optional, edition parameter is omitted, String.isIdentifierStart and String.isIdentifierPart are supposed to return true if the current engine supports it as such, and false otherwise (as per the proposal in the strawman). Because this is a polyfill for use in older ECMAScript environments, I’ve decided to use the ECMAScript 6 definition for identifiers and the latest available Unicode version data to determine the result instead. In a proper ES6 environment, String.isIdentifierPart(0x2FA1D) is always true — without violating the strawman, this polyfill could never return that result in a non-ES6 environment, which didn’t seem very useful.

Author

| twitter/mathias | |---| | Mathias Bynens |

License

This polyfill is available under the MIT license.