how to use local variable in other functions

Question

This is my code:

var shuffle = function(x) {
    var deck = [];
    deck.push(x);

    return deck;
};

var Placemat = function() {
    var card1 = deck.shift();
    var card2 = deck.shift();
}

By returning deck in shuffle() like this, can I use it in Placemat()? If not, how can I do it?

I don't know about that username bro. – Kenan Banks Aug 10 '11 at 20:28 — Kenan Banks, Aug 10 '11 at 20:28
@Triptych--fair enough, but do you know the answer? – dopatraman Aug 10 '11 at 20:30 — dopatraman, Aug 10 '11 at 20:30

jAndy · Accepted Answer · 2011-08-10T20:37:33.957

Yes you could. But anytime you call .shuffle() you would overwrite that array, or more precisely, create a new instance. So if you want to keep that reference you could go like

var shuffle = (function() {
    var deck = [];

    return function(x) {
        deck.push(x);

        return deck;
    };
}());

Now .shuffle() closes over the deck by returning another function. So it could look like

var Placemat = function() {
    var myDeck = shuffle(5);

    shuffle(10);
    shuffle(15);

    var card1 = deck.shift();
    var card2 = deck.shift();
}

Even if that is probably not the greatest way to go. But I guess I don't even know what exactly you want to achieve there.

Shouldn't it be `myDeck.shift()` ? – Jiri Kriz Aug 10 '11 at 20:49 — Jiri Kriz, Aug 10 '11 at 20:49

score 0 · Answer 2 · answered Aug 10 '11 at 20:29

You cannot you have to put them in the parameters or outside the function.

var shuffle = function(x) {
    var deck = [];
    deck.push(x);

    return deck;
};

 var Placemat = function(deck) {
    var card1 = deck.shift();
    var card2 = deck.shift();
}

or 

var deck = [];
var Placemat = function() {
    var card1 = deck.shift();
    car card2 = deck.shift();
}

score 0 · Answer 3 · answered Aug 10 '11 at 20:31

You can use the return value from shuffle in Placemat, but it will always be an array with a single item in it.

If you want to have a variable deck that contains multiple items, deck will need to be declared in an outer scope:

var deck = [];

var shuffle = function(x) {
    deck.push(x);
    return deck;
}

var Placemat = function() {
    var card1 = deck.shift();
    var card2 = deck.shift();
};

This way, both the shuffle function and Placemat are using the same variable deck.

score -1 · Answer 4 · answered Aug 10 '11 at 20:53

I found another solution:

var shuffle = function(deck,x) {
    var deck = [];
    deck.push(x);

    return deck;
};

var Placemat = function() {
    var newdeck = shuffle(deck,x)
    var card1 = newdeck.shift();
    var card2 = newdeck.shift();
}

This way I don't have to make deck a global variable. Thanks for the help everyone.

how to use local variable in other functions

4 Answers4