Trivial Inheritance with JavaScript

Question

function StringStream() {}
StringStream.prototype = new Array();
StringStream.prototype.toString = function(){ return this.join(''); };

Calling new StringStream(1,2,3) gives an empty array

x = new StringStream(1,2,3)

gives

StringStream[0]
__proto__: Array[0]

Can someone please explain why the superclass' (Array) constructor is not called?

who gave you this code and why ? :(

c69
– c69

2011-10-08 15:56:53 +00:00
Commented Oct 8, 2011 at 15:56 — c69
– c69, Commented Oct 8, 2011 at 15:56

pimvdb · Accepted Answer · 2011-10-08 15:57:58Z

2

Just because StringStream.prototype is an array, the StringStream constructor is not replaced with Array as well.

You should implement that yourself: http://jsfiddle.net/gBrtf/.

function StringStream() {
    // push arguments as elements to this instance
    Array.prototype.push.apply(this, arguments);
}

StringStream.prototype = new Array;

StringStream.prototype.toString = function(){
    return this.join('');
};

answered Oct 8, 2011 at 15:57

pimvdb

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2 Comments

Markos Evlogimenos Over a year ago

So there is no way to call the constructor of Array obj and passing the arguments instead of invoking the push method?

pimvdb Over a year ago

@Markos Evlogimenos: You cannot combine new and .apply so I guess there is no way. Anyway, why are you trying to create a constructor that is equivalent to Array?

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