Consistency of contains method on a HashSet and HashMap

Question

Why does containsAll method on a HashSet does not remain consistent if remove is called on the Set whereas a containsValue method on a HashMap remains consistent after a value is removed After a value is removed from a HashSet containsAll returns false even if all values were present where as in case of HashMap the containsValue method returns correct value

public static void main(String[] args)
    {

    HashSet<String> lookup=new HashSet<String>();
    HashMap<Integer,String> findup=new HashMap<Integer,String>();

    String[] Alltokens={"This","is","a","programming","test","This","is","a","any","language"};
    for(String s:Alltokens)
        lookup.add(s);

    String[] tokens={"This","is","a"};
    Collection<String> tok=Arrays.asList(tokens);

    lookup.remove(Alltokens[5]); 
     if(lookup.containsAll(tok))
       System.out.print("found");
    else    
        System.out.print("NOT found");

    Integer i=0;
    for(String s:Alltokens)
        findup.put(i++,s);
    boolean found=true;
    findup.remove(Alltokens[0]);
        findup.remove(5);               //Edit : trying to remove value whose key is 5
    for(String s:tokens)
        if(!findup.containsValue(s))
            found=false;
    System.out.print("\nIn map " + found);
}

Output NOT found In map true

Is there a way to keep containsAll consistent if remove method is called on the HashSet? Also if a value that was not present in the set is passed to remove method.ContainsAll remains consistent

        lookup.remove("Alltokens[5]");
    if(lookup.containsAll(tok))

//This will be true now where as it is false if a value already present is removed

May be it has got to do something with keys in HashMaps and no keys in HashSet.Can you please explain how do they work?

There's something missing after `lookup.remove(Alltokens[5]);`. — Joachim Sauer, Nov 12 '12 at 07:54
The above code will not compile. You should post the actual code, and you should add `{ }` rather than allowing one-line statements to cause these issues for you, as you refactor your code and they change from one-liners to multi-liners. — pickypg, Nov 12 '12 at 07:55
@JoachimSauer Sorry the if condition went missing.edited it now — rakesh99, Nov 12 '12 at 08:18

score 5 · Accepted Answer · answered Nov 12 '12 at 07:55

5

Map.remove(Object) removes a mapping based on the key.

Since you use Integer objects as the keys (put(i++, s)) you will remove nothing when you call findup.remove(Alltokens[0])! Check its return value to see that it will return false.

answered Nov 12 '12 at 07:55

Joachim Sauer

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I changed it to findup.remove(5) even then it returns true.Moreover how would this figured out if i have passed a key or a value.If the method takes a key as an argument won't it have tried to convert a string to an integer in the case findup.remove(Alltokens[0]) – rakesh99 Nov 12 '12 at 08:05
1

1.) 5="This" and "This" is contained twice in your map: once at 0, once at 5. So even after you removed 5, "This" is still contained in the values of the map and 2.) No, a `String` is simply never `equal` to an `Integer`, so it will just never remove anything. Java doesn't do automagic type conversion for you, most of the time. – Joachim Sauer Nov 12 '12 at 08:12
And in case of a HashSet after Alltokens[5] is removed since it doesn't contain duplicates 'This' is no more there and it returns false.But why doesn't findup.remove(Alltokens[0]) give an error.Isn't it an error to pass a string to a function which takes an integer? – rakesh99 Nov 12 '12 at 08:24
1

The reason that compiles is that `remove()` accepts an `Object`, no matter what the key type is. The [reason is complicated](http://stackoverflow.com/questions/857420/what-are-the-reasons-why-map-getobject-key-is-not-fully-generic) and *most of the time* it's an error to pass in something that's not of the type of your key, but unfortunately that decision is fixed and can't easily be changed. – Joachim Sauer Nov 12 '12 at 08:30

Consistency of contains method on a HashSet and HashMap

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