CODEWITHSUNDEEP
iosswift
Karthik Menon
Karthik Menon

Reputation: 71

Swift: Dictionary passed as parameter is always empty

I want to fill Dictionary via a method of which dictionary is one of the parameter. When I add a value to key, say 15, it always return 0 count when I try to access it second time with same key i.e. 15. Here's the code.

private static var showWidgetMap = Dictionary<Int, [BaseWidget]>() private static var hideWidgetMap = Dictionary<Int, [BaseWidget]>() 

static func initHandler()
{
    let cloudWidget = CloudWidget()
    cloudWidget.setType(CreatorConstants.CLOUD)
    let property1 = [CreatorConstants.IMG_SRC: "cloud1", CreatorConstants.X_COORD: "100", CreatorConstants.Y_COORD: "450"]
    cloudWidget.setPropery(property1)

    addWidgetInLocalTimeList(15, widget: cloudWidget, delete: false)

    let emojiWidget = CloudWidget()
    emojiWidget.setType(CreatorConstants.EMOTICON)
    let property2 = [CreatorConstants.IMG_SRC: "1", CreatorConstants.X_COORD: "100", CreatorConstants.Y_COORD: "550"]
    emojiWidget.setPropery(property2)

    addWidgetInLocalTimeList(15, widget: emojiWidget, delete: false)}

static func addWidgetInLocalTimeList(time_milisec: Int, widget: BaseWidget, delete: Bool) { if(delete) { checkAndAdd(hideWidgetMap, key: time_milisec, widget: widget); } else { checkAndAdd(showWidgetMap, key: time_milisec, widget: widget); } } 

private static func checkAndAdd(var map: Dictionary<Int, [BaseWidget]>, key: Int, widget: BaseWidget)
{
    print("map count is")
    print(map.count)

    if var val = map[key]
    {
        val.append(widget);
    }
    else
    {
        var temp: [BaseWidget]  = [];
        temp.append(widget);
        map[key] = temp
        print(map.count)
    }
}

print(map.count) always returns 0.

Upvotes: 0

Views: 307

Answers (1)

Sweeper
Sweeper

Reputation: 271040

You need to understand the difference between value types and reference types.

Value type variables are just values. For example, an array is a value type. It is just a value of "a bunch of stuff" *. On the other hand, reference types are references to values. For example, when you create a UIViewController, that variable actually stores a reference to the actual UIviewController *.

Don't really understand? Then it's analogy time! The variables and constants you create are children. The things you put in variables and constants are balloons.

There are two types of children, one type (value types) likes to hold balloons directly in their hands. The other type (reference types) likes to hold balloons using a string **.

When you pass a child to a method, depending on what type of child he is, different things will happen:

  • A value type child holds the balloon in his hands, so tightly that the method parameter can't take it away from him. So what can it do? It creates a copy of it! It then takes the copy to the method implementation let it do its thing.

  • A reference type, however, holds balloons using a string. The method parameter will tie another string to the balloon so the implementation can access it using the string. As a result, no copies of the balloon are created.

So what are you doing wrong here?

Since swift dictionaries are value types, when you pass a dictionary to a method, as I said above, it creates a copy! In the implementation, you are actually editing a copy of the dictionary, not the original one. That's why the original dictionary still has a count of 0.

What can you do?

Instead of marking the parameter with var, which is a very bad practice btw, you mark it with inout!

private static func checkAndAdd(inout map: Dictionary<Int, [BaseWidget]>, key: Int, widget: BaseWidget)

The inout modifier basically says

Hey parameter, next time you see a value type, just get a string and tie it to the balloon that the child is holding.

There is also another thing that you should do. That is you should change the way you call your method.

Instead of 

checkAndAdd(showWidgetMap, key: time_milisec, widget: widget)

You write

checkAndAdd(&showWidgetMap, key: time_milisec, widget: widget)

And magically, it works!

Conclusion: Parameters are dumb. They aren't even smart enough to know when to tie a string. Be careful when you work with value types.

Footnotes * Assume it is not nil

** Not the String type, but an actual string.

Upvotes: 1

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