What does the bitwise shift left accomplish here?

Question

https://stackoverflow.com/a/15106541/10757573

In this thread, the solution uses the following code:

var randomProperty = function (obj) {
    var keys = Object.keys(obj)
    return obj[keys[ keys.length * Math.random() << 0]];
};

In the comments from the answer, one user explained the bitwise shift as:

It's more as a shorthand of parseInt(keys.length * Math.random(), 0)

Can anyone explain how the bitwise shift left 0 is the same as parseInt, or how it works in general, please?

I'm using this syntax, and it works well, but I don't understand what it's doing.

"*I'm using this syntax, and it works well, but I don't understand what it's doing.*" then how do you know it works *well*? — VLAZ, Sep 12 '19 at 05:06
Also, `parseInt(keys.length * Math.random(), 0)` makes no sense - you parse the value into base zero? What is any number in base zero represented as? — VLAZ, Sep 12 '19 at 05:08
@VLAZ `radix: 0` actually has special behaviour: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt — Dai, Sep 12 '19 at 05:09

score 2 · Answer 1 · answered Sep 12 '19 at 05:14

The usual idiom for selecting a random array element is

array[Math.floor(array.length * Math.random())]

What this code is doing is using << 0 as a shortcut for Math.floor(). keys.length * Math.random() will be a floating point number. The << operator requires that its operands be integers, so it will automatically convert that float to an integer, and it does this by rounding down, just like Math.floor() does.

score 2 · Accepted Answer · answered Sep 12 '19 at 05:14

Performing value << 0 in JavaScript has the effect of coercing value into an integer when value is a non-integer. This is needed in this case because Math.random() returns a floating-point number, which cannot be used as an integer indexer (keys is an Array so it's indexed by an integer).

parseInt( /*string:*/ textValue, /*radix:*/ 0 ) does not convert a number to "base 0", instead when radix == 0, the parseInt function will parse textValue as base 10 unless textValue starts with 0x in which case it will parse it as base 16.

What does the bitwise shift left accomplish here?

2 Answers2