Find an Array Value using Two Variables

Question

Suppose I set up a simple array:

var list = [
    "Hello",
    "Goodbye",
];

And then I set two variables, one to specify two variables: one for the name of the array, and another for the number, as such:

var array = "list";
var number = 0;

And then I run the following command:

console.log(array[number]);

But it outputs the number "1" instead of "Hello".

Can somebody explain this?

EDIT:

I wrote this question as an inexperienced programmer. What I should have used to complete the task was a dictionary.

They will allow you to, very roughly speaking, index arrays with strings.

You must referance the current scope. If that is global then you would use console.log(window[array][number]) as window is the global scope. — Blindman67
– Blindman67, Commented Jun 26, 2016 at 4:34
@Blindman67 That can be done, but not recommended. And, from the variable name in var array = "list"; you can get idea what OP want to do here. — Tushar
– Tushar, Commented Jun 26, 2016 at 4:36
@Tushar I would guess he is trying to indirectly referance the array list. Using bracket notation is a legitimate method to access variables, this[array][number] would have worked as well. — Blindman67
– Blindman67, Commented Jun 26, 2016 at 4:43

Tushar · Accepted Answer · 2016-06-26 04:38:44Z

5

But it outputs the number "1" instead of "Hello". Any ideas?

No, it doesn't output 1, it's l(the alphabet L lowercased).

array is no longer an array when you assign a string list to it in

var array = "list";

and then

array[0]

is equivalent of

"list"[0]

which will return the character at zeroth index, thus returning l.

Solution:

To reference an array, use the variable name(no quotes required)

var array = list; // Remove quotes.

If you don't want to reference same array and want to create a copy, use slice()

var copy = list.slice(0);

edited Jun 26, 2016 at 4:38

answered Jun 26, 2016 at 4:27

Tushar

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Comments

pgreen2 · Accepted Answer · 2016-06-26 04:40:29Z

Are you sure it is outputting "1" (number one) and not "l" (lowecase letter ell)? Because

console.log(array[number]);

starts with the array object (which is a string of value 'list') and then returns the position associated with number, which is 0. So in english this means print the 0th letter of 'list' which is 'l'.

Based upon what you wrote, is sounds like you are being very abstract with a level of indirection I don't normally see. You can do what you are asking, but it is a strange thing to do. list is a variable on the global scope, which for a browser is available through the window. So you could do the following:

window[array][number]

which is interpreted to the following:

window['list'][number]

which is the same as

list[number]

which is the same as

list[0]

which is

"Hello"

Generally, the level of abstraction you specify with the array variable is unnecessary.

Tad Donaghe · Accepted Answer · 2016-06-26 04:27:52Z

1

console.log(array[number])

is printing out the value of the 0th element of the string "list" which is l, which is what you're seeing.

You have assigned the string "list" the a variable named array. And then when you attempt to retrieve the 0th (you variable number) element, you get the first character in the string "list".

answered Jun 26, 2016 at 4:27

Tad Donaghe

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