Initialize array with variables: Unexpected token +

Question

I created a simple function which should initialize an array and return one random element from it everytime when the function is called.

function test(str)
{
    var myArray = 
    {
        "Test1 "+ str + " Test1",
        "Test2 "+ str+ " Test2"
    }

    return myArray[Math.random()*myArray.length+0];
}

console.log(test("FOO"));

But I get Uncaught SyntaxError: Unexpected token +

`var myArray = [ "Test1 "+ str + " Test1", "Test2 "+ str+ " Test2" ]` — Rayon, Apr 22 '16 at 11:51
return myArray[ Math.floor((Math.random() * myArray.length)) ]; — user5200704, Apr 22 '16 at 12:35

Aboelseoud · Answer 1 · 2016-04-22T11:56:40.280

2

Use [ ] instead of { } for a JS Array.

function test(str)
    {
        var myArray =
                [
                    "Test1 "+ str + " Test1",
                    "Test2 "+ str+ " Test2"
                ]

        return myArray[parseInt(Math.random()*myArray.length, 10)];
    }

edited Apr 22 '16 at 11:56

answered Apr 22 '16 at 11:52

Aboelseoud

556
1
7
22

2

@upvoters This will give `undefined` all the time. – Rajaprabhu Aravindasamy Apr 22 '16 at 11:54
@RajaprabhuAravindasamy Updated to parse double to int. – Aboelseoud Apr 22 '16 at 11:56
Thank you, but i think the solution from @RajaprabhuAravindasamy is easier to understand. – Black Apr 22 '16 at 12:01
1

@EdwardBlack parseInt(str, b) parses str to an integer in base b. – Aboelseoud Apr 22 '16 at 12:04
@EdwardBlack Decimal. – Aboelseoud Apr 22 '16 at 12:04
1

-1. [Do not use `parseInt` on numbers](http://stackoverflow.com/q/32039288/1048572)! – Bergi Apr 22 '16 at 12:07

Rajaprabhu Aravindasamy · Accepted Answer · 2016-04-22T12:13:25.227

2

Here in your code myArray is not an array,

function test(str){
    var myArray = ["Test1 " + str + " Test1","Test2 " + str + " Test2"] 
    return myArray[(Math.random() * myArray.length) | 0];
}

| This is bit wise OR operator. When doing operand1 | operand2, both the operands will get converted to base 2 and performs the OR operation over it. Finally the result will get converted back to the base 10. Hence the decimal points will be removed.

edited Apr 22 '16 at 12:13

answered Apr 22 '16 at 11:52

Rajaprabhu Aravindasamy

66,513
17
101
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This is the best solution i think. – Black Apr 22 '16 at 12:01
@EdwardBlack Glad to help! :) – Rajaprabhu Aravindasamy Apr 22 '16 at 12:03
Yes! Nice solution.. Just explain `|` :) – Rayon Apr 22 '16 at 12:03
"*converted to base 2 that is binary*" is confusing, if not wrong. Yes, the number will be converted to a 32 bit integer, but everything on a computer is "binary". And there is no "*original base*", it's a floating point number. – Bergi Apr 22 '16 at 12:10
@Bergi Corrected the things that you said. But base 2 numbers cannot be told as binary? I have a confusion on this. – Rajaprabhu Aravindasamy Apr 22 '16 at 12:14
Oh, sure "binary" can mean the same as "base 2", just like "decimal" stands for "base 10". But every number on a computer really already is in a binary format - unless it is stored in a string - so this "base conversion" is meaningless. What really happens is a conversion from floating point to integer format, which strips the decimal digits. Or just use `Math.floor`, then you don't have to [explain](http://stackoverflow.com/q/9049677/1048572) [anything](http://stackoverflow.com/q/7487977/1048572). – Bergi Apr 22 '16 at 12:27

Initialize array with variables: Unexpected token +

2 Answers2