32-bit signed integer math in JavaScript

Question

I am converting some legacy Pascal to JavaScript. I need to multiple two 32-bit signed integers.

In the following sample loop some multiplications will cause overflow and will give negative numbers. This is intentional. I need to reproduce the same final number x at the end that matches the legacy system.

How can I do this in JavaScript to achieve the same result?

Here is some sample code:

var x = new Number(some value);  // I need this to be a 32-bit signed integer
var y = new Number(some value); // I need this to be a 32-bit signed integer

for (var i=0; i<100; i++) {   
    x = x * y; 
} 
return x;

Adam D. Ruppe · Accepted Answer · 2013-11-07 18:47:59Z

10

Javascript's bitwise operators actually convert the value to a regular integer. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators This fact is used by things like asm.js to coerce the types, and you can do it yourself too. The trick is to put a |0 at the end of a number to force it to be 32 bit

function test() {
    var x = 255|0; // |0 does the type coercion
    var y = 255|0; // not strictly necessary at this var decl but used for explicitness

    for (var i=0; i<5; i++) {   
         x = (y * x)|0; // parens needed because |'s precedence
   } 

   return x;
}

I ran that with a few numbers and got the same result as C in Firefox.. didn't get a chance to test in IE, but I'm pretty sure this behavior is in the ECMAscript spec, so it should work.

edited Nov 7, 2013 at 18:47

answered Nov 7, 2013 at 18:37

Adam D. Ruppe

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4 Comments

XCS Over a year ago

I don't think the parnes are required.

Adam D. Ruppe Over a year ago

| has higher precedence than '*'. But my rule of thumb anyway is if you have to look up or even think more than a second about precedence rules, just write the parens to make is double-clear for the next reader anyway.

XCS Over a year ago

The | operator has a very low precedence, * has a very high precedence and is most likely to be executed first in operations. See this: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Adam D. Ruppe Over a year ago

huh I should have looked it up... you are indeed right. But still, I prefer to write them anyway.

Pointy · Accepted Answer · 2013-11-07 18:25:11Z

1

Math operations in JavaScript are always done as double-precision floating point. You'd have to write your own multiplication routine (or find one somewhere) to carry out integer math, and that'd be slow or hard (or both :).

answered Nov 7, 2013 at 18:25

Pointy

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7 Comments

Pointy Over a year ago

This is one of the most critically weak aspects of the language, by the way.

Vic Over a year ago

What you say is true but the accepted answer above solves my current problem.

Pointy Over a year ago

@Vic well it'll work for some values, but the multiplication operation itself is carried out with floating point hardware as a double-precision operation, so values near the limit of numeric capacity may get weird.

Pointy Over a year ago

@Vic actually since there are plenty more mantissa bits than 32, it may work out OK.

Vic Over a year ago

So far my test data has shown the right results. Keeping my figures crossed.

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