Prototype vs Module pattern performance

JavaScript performance comparison

Revision 54 of this test case created by Aaron

Info

Removed iterations in tests, because jsperf already does that for us. And renamed variables to be more meaningful to us humans. And other cosmetic changes.

The most important thing to remember is to use the right tool for the job. All these tests do is reference an object with a complex memory allocation. When you don't need something fancy, you're better off using a regular old object

Preparation code

 
<script>
  Benchmark.prototype.setup = function() {
    function TraditionalPrototypeClass() {
    }
    
    TraditionalPrototypeClass.prototype.foo = function() {
    };
    
    TraditionalPrototypeClass.prototype.bar = function() {
    };
    
    function ModulePatternClass() {
        this.foo = function() {
        };
        
        this.bar = function() {
        };
    }
    
    var ModuleCachePatternClass = (function () {
        function foo() {
        }
        
        function bar() {
        }
        
        return function () {
            this.foo = foo;
            this.bar = bar;
        };
    }());
    
    var standardObject = {
        foo: function(){
        },
        bar: function(){
        }
    };
  };
</script>

Test runner

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Testing in unknown unknown
Test Ops/sec
Prototypal
 
var o = new TraditionalPrototypeClass()
o.bar;
o.foo;
 pending…
Module pattern
 
var o = new ModulePatternClass()
o.bar;
o.foo;
 pending…
Module pattern with cached functions
 
var o = new ModuleCachePatternClass()
o.bar;
o.foo;
 pending…
Use the right tool for the job
 
var o = standardObject;
o.bar;
o.foo;
 pending…

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Revisions

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1 comment

Montana Harkin commented : ∞

Your "right tool" test case isn't actually creating a new object. See my revision.

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