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Array.concat() and it’s uses…

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Sep 11.2007

I’m just wondering if anyone can help me with an issue:

Say I have either of the examples below:

var fruits = new Array(“lemon”, “apple”, “orange”, “lime”)

or

var fruits = new Array()
fruits[0] = “lemon”;
fruits[1] = “apple”;
fruits[2] = “orange”;
fruits[3] = “lime”;

Now, why, if I use the concat(), do I receive an “unidentified” attached to the end?

fruits[0].concat(fruits[3]);

I’m looking for “lemonlime”, not “lemonlimeunidentified”…

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JavaScript

5 Comments(s) ↴

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@mrhooSep 11.2007 — var fruits =["lemon", "apple", "orange", "lime"];

alert(fruits[0].concat(fruits[3]));

I have no problem with this in IE7,Opera9,or Firefox2-

I had to check, I always use fruits[0][B]+[/B] fruits[3].
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@WebTechauthorSep 11.2007 — mrhoo

did you try the second example to see if it would work?

I was using:

document.write(fruits[0].concat(fruits[3]);
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@mrhooSep 11.2007 — var fruits = new Array()

fruits[0] = "lemon";

fruits[1] = "apple";

fruits[2] = "orange";

fruits[3] = "lime";

alert(fruits[0].concat(fruits[3]));

I get lemonlime for the alert and for document.write...
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@WebTechauthorSep 11.2007 — try the same with "document.write();". maybe that's where I'm having the problem...
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@mrhooSep 12.2007 — This has nothing to do with your problem, which I can't reproduce, but your title is wrong. You are using String.concat, not Array.concat.

Array.concat returns a new Array by chaining Arrays together, end to end.

var A=[1,2,3];

var B=[3,2,1];

var C= A.concat(?;

A and B still have their original values,

C=[1,2,3,3,2,1];

String has its own concat, which you are using on individual Array members.

Same as using '+'.
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