CODEWITHSUNDEEP
javascriptoptimizationbinarybit-manipulationbitwise-operators
user5124106
user5124106

Reputation: 403

What is the flaw in my bitwise logic?

What I'm trying to do at the beginning of my function is easy to see from my comment

    this.move = function ( )
    {

        if (this.dx | this.dy != 0) return; // exit early if both this.dx and this.dy are zero

That is, I want the equivalent of if (this.dx == 0 && this.dy == 0). I thought the bitwise OR is correct because this.dx | this.dy is not equal to zero if and only if this.dx has at least one bit on or this.dy has at least one bit on (or both have at least one bit on). But I must be wrong, because my test

    this.move = function ( )
    {
        console.log("this.dx = " + this.dx + ", this.dy = " + this.dy); // TEST

        if (this.dx | this.dy != 0) return; // exit early if both this.dx and this.dy are zero

is showing that the rest of the function is executing when this.dx and this.dy are both zero.

What's going on here?

Upvotes: 1

Views: 40

Answers (2)

musically_ut
musically_ut

Reputation: 34288

According to the precedence table The bitwise OR will be executed after the inequality check is done. For example:

[JS]> 0 | 0 == 0
1

Hence, your expression actually is executed as:

if (this.dx | (this.dy != 0)) { ... }

To fix the issue, parenthesize the bitwise OR: if ((this.dx | this.dy) != 0).

Also, as @Jon-Skeet pointed out, the correct check probably should be with an ==.

Upvotes: 1

user2575725
user2575725

Reputation:

Problem is the condition is executed as:

this.dx | (this.dy != 0)

Try this:

if (!(this.dx | this.dy)) return;

Upvotes: 3

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