Cannot overwrite "this" in javascript when extending String with String.prototype

Question

I want to create a mutable reverse method for a javascript string.

TL; DR:

Here is my attempt that did not work:

String.prototype.reverse = function() {


    var reversed = {};
    j = 0;
    for (i = this.length - 1; i >= 0; i--)
    {
        reversed[j++] = this[i];
    }
    for (i in reversed)
    {
        this[i] = reversed[i];
    }
};

...

str1 = "hello";
str1.reverse();
console.log(str1); //hello, not olleh

It doesn't change the string at all, as further evidenced by this small test:

String.prototype.makeFirstCharX = function() {
    console.log(this[0]); //h
    this[0] = 'x'; //no error
    console.log(this[0]); //h ??
};

str1.makeFirstCharX();
console.log(str1); //hello, not xello

Is there anyway to overwrite values of this in a new protoype function for a native JS type? Why doesn't something like this[0]='x' give an error instead of just silently failing?

---This is what I tried on a custom JS object that did work as expected, which is what I based the above off of:

I can create a custom revsersible string that behaves similarly to a native string like this:

    function MyString(str) {
//set string content and length
        var l = 0;
        for (i in str)
        {
            this[i] = str[i];
            l++;
        }
        this.length = l;

    }
    MyString.prototype.toString = function()
    {
        var retVal = '';
        for (i = 0; i < this.length; i++)
        {
            retVal += this[i];
        }
        return retVal;
    };

    MyString.prototype.reverse = function()
    {
        var reversed = {};
        j = 0;
        for (i = this.length - 1; i >= 0; i--)
        {
            reversed[j++] = this[i];
        }
        for (i in reversed)
        {
            this[i] = reversed[i];
        }
    };

So as expected:

    str = new MyString('string');
    console.log(str); // MyString { 0="s", 1="t", 2="r", more...}
    console.log("" + str); //string
    console.log(str.length); //6
    str.reverse();
    console.log(str); // MyString { 0="g", 1="n", 2="i", more...}
    console.log("" + str); //gnirts

But if I try the same thing on a "real" javascript string, it does not work.

Short answer: Strings in javascript are immutables

Aguardientico
– Aguardientico

2014-07-28 00:01:19 +00:00
Commented Jul 28, 2014 at 0:01 — Aguardientico
– Aguardientico, Commented Jul 28, 2014 at 0:01

G-- · Accepted Answer · 2014-07-28 00:07:31Z

First Strings aren't mutable in Javascript.

from the rhino book:

In JavaScript, strings are immutable objects, which means that the characters within them may not be changed and that any operations on strings actually create new strings. Strings are assigned by reference, not by value. In general, when an object is assigned by reference, a change made to the object through one reference will be visible through all other references to the object. Because strings cannot be changed, however, you can have multiple references to a string object and not worry that the string value will change without your knowing it

Also Javascript passes it by value. Which means that you aren't modifying your original string anywhere. You have to return something.

Here's a simple implementation for the reverse function:

function reverse(s) {
  var reversed = '';
  for (var i = s.length - 1; i >= 0; i--)
    reversed += s[i];
  return reversed;
}

You have to reassign the original var.

var original = 'original';
original = reverse(original);

fgb · Accepted Answer · 2014-07-28 00:04:21Z

1

The bracket notation isn't going to work. From String:

For character access using bracket notation, attempting to delete or assign a value to these properties will not succeed. The properties involved are neither writable nor configurable. (See Object.defineProperty for more information.)

Strings are immutable in Javascript so you're not going to be able to change them by any means.

answered Jul 28, 2014 at 0:04

fgb

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1 Comment

chiliNUT Over a year ago

Thanks, this at least explains why overwriting this will silently fail.

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