Scope of variables in JavaScript, function does not return correct value

Question

Good day, I've stumbled upon a trouble with scopes of variables in JS. I have a function like this

var address = ...; //it may be or may not be defined earlier.
// It is not the only variable passed to the function

function search() {
    window.location.search = address;
}

But I need to make sure this variable in defined, so i do this

function search() {
    function check(value) {
        if(!value) {
            value = "1";
        }
        return value;
    }
    check(address);
    window.location.search = address;
}

But what I get sometimes is undefined,passed to the window.location.search, since the check() does return the "1" properly;

Can you please point out my mistake?

[UPDATE]

I should have said earlier - im just looking for a better way to do it. Sure simple if check will get it right, but the $address variable is not the only one You can picture it like this

totaladdress = address1;
totaladdress += address2;
totaladdress += address3;
totaladdress += address4;
totaladdress += address5;
...
window.location.search = totaladdress;

So I need a function to check them (e.g. check(address1);) and avoid several dozens of IFs to make the code cleaner.

If it may or may not be defined you need to use typeof foo === 'undefined'. — elclanrs
– elclanrs, Commented May 16, 2013 at 7:13
Did you want window.location.search = check(address);? You don't use the result of your check call anywhere — Bergi
– Bergi, Commented May 16, 2013 at 7:43

tom · Accepted Answer · 2013-05-16 07:42:26Z

1

When you assign "1" to value it does not modify address because the String reference associated with address is copied into value, and you cannot modify the original reference. Example:

var address = "abc";
var value = address;
value = "def";
// address is still "abc"

You correctly return the string from the function, but the returned value is ignored so it doesn't help. What you want is

address = check(address);
window.location.search = address;

or just

window.location.search = check(address);

or even

function setAddress(value) {
    window.location.search = value || "1";
}

answered May 16, 2013 at 7:42

tom

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Comments

ThiefMaster · Accepted Answer · 2013-05-16 07:14:35Z

1

You cannot use a function for this kind of check if the variable may be undefined - you need to use typeof directly at the place where you use the possibly undefined name:

function search() {
    window.location.search = typeof address !== 'undefined' ? address : 'some default value';
}

answered May 16, 2013 at 7:14

ThiefMaster

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1 Comment

Kresent Over a year ago

That is a vaild option, but the thing is - if i log the variable from within check function - it is correctly displayed as "1" if undefined. The problem is that this "1" is discarded after that and I get undefined again in the end.

Arun P Johny · Accepted Answer · 2013-05-16 07:14:46Z

0

You re not returning any value from the check method.

function search() {
    function check(value) {
        if(!value) {
            value = "1";
        }
        return value;
    }
    window.location.search = check(address);
}

answered May 16, 2013 at 7:14

Arun P Johny

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Comments

HMR · Accepted Answer · 2013-05-16 07:15:27Z

0

[UPDATE]Sorry misunderstood what you needed. You can try this.

var address = ...; //it may be or may not be defined earlier.
// It is not the only variable passed to the function

function search(){
  window.address = (window.address === undefined)? 1:window.address;
  window.location.search = window.address;
}

answered May 16, 2013 at 7:15

HMR

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