Why does JSHint argue against bitwise operators? How should I express this code?

javascript jslint jshint
Sean Anderson picture Sean Anderson · Jul 23, 2012 · Viewed 21.1k times · Source

I am using this bit of JavaScript to generate a UID:

(original:)

//If ID has not been defined then generate a new unique ID.
if(!id){
    id = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) { var r = Math.random() * 16 | 0, v = c == 'x' ? r : (r & 0x3 | 0x8); return v.toString(16); });
}

(formatted so it can be read:)

// If ID has not been defined then generate a new unique ID.
if (!id) {
    id = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(
        /[xy]/g, 
        function (c) { 
            var r = Math.random() * 16 | 0, 
                v = c == 'x' ? r : (r & 0x3 | 0x8); 
            return v.toString(16); 
        }
    );
}

JSHint does not like the use of bitwise OR and AND operators. I was wondering how I could rewrite this to be more 'standard friendly.'

EDIT: JSHint states:

Line 8: id = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) { var r = Math.random() * 16 | 0, v = c == 'x' ? r : (r & 0x3 | 0x8); return v.toString(16); });
Unexpected use of '|'.

Line 8: id = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) { var r = Math.random() * 16 | 0, v = c == 'x' ? r : (r & 0x3 | 0x8); return v.toString(16); });
Expected '===' and instead saw '=='.

Line 8: id = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) { var r = Math.random() * 16 | 0, v = c == 'x' ? r : (r & 0x3 | 0x8); return v.toString(16); });
Unexpected use of '&'.

Line 8: id = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) { var r = Math.random() * 16 | 0, v = c == 'x' ? r : (r & 0x3 | 0x8); return v.toString(16); });
Unexpected use of '|'.

Answer

zerkms picture zerkms · Jul 23, 2012

Put

/*jshint bitwise: false*/

in the top of your file

A list of available options: http://jshint.com/docs/options/

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