Why does << 0 remove the decimal portion of a number in JavaScript?

Question

When you have a floating-point number, say 10.1, and you utilize the zero fill left shift operator (<<) in JavaScript, it returns just the integer portion of the number, say 10 for the example.

Here's some source code showing what I mean:

var num1 = 958.51243

console.log(num1) // 958.51243
console.log(num1 << 0) // 958

Why does this happen?

come on. ecma-262 is a standard. the language follows the standard. why ... is always a question of the standard. — Nina Scholz, Jan 18 '20 at 22:07
From the standard: *The result is a signed 32-bit integer.*. And here's the standard which wasn't hard to find: https://www.ecma-international.org/ecma-262/5.1/#sec-11.7.1 — GetSet, Jan 18 '20 at 22:10

score 1 · Accepted Answer · edited Jun 20 '20 at 09:12

1

See the docs:

The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format, except for zero-fill right shift which results in an unsigned 32-bit integer.

So the numbers will be converted to integers before the operation is performed.

For left shift, it's described in the official specification here:

6.1.6.1.9 Number::leftShift ( x, y )

Let lnum be ! ToInt32(x).

Let rnum be ! ToUint32(y).

(...perform operation...)

Where ToInt32 does:

The abstract operation ToInt32 converts argument to one of 2^32 integer values in the range -2^31 through 2^31 - 1, inclusive.

edited Jun 20 '20 at 09:12

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answered Jan 18 '20 at 22:03

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Awesome, thanks for the descriptive answer! – Hays Stanford Jan 19 '20 at 02:55
I cant think of any reason why i would want to bitwise shift a floating point number – GetSet Jan 20 '20 at 00:57
To convert it to an integer. – Hays Stanford Jan 23 '20 at 20:58

Why does << 0 remove the decimal portion of a number in JavaScript?

1 Answers1