CODEWITHSUNDEEP
javacollectionssetcontains
David Pérez Cabrera
David Pérez Cabrera

Reputation: 5068

Why containsAll does return true in second case

It compiles and runs without problems, but I cannot understand why second statement return false: 

    String s0 ="0123";
    String s1 ="0123";


    System.out.println(new HashSet<>(Arrays.asList(ArrayUtils.toObject(s1.toCharArray()))).containsAll(Arrays.asList(ArrayUtils.toObject(s0.toCharArray()))));
    System.out.println(new HashSet<>(Arrays.asList(s1.toCharArray())).containsAll(Arrays.asList(s0.toCharArray())));

Output:

true
false

JDK 1.8

EDIT:

Thank you @Eran There is a compilation problem when I avoid diamond operator like this:

new HashSet<Character>(Arrays.asList(s1.toCharArray())).containsAll(Arrays.asList(s0.toCharArray()));

Upvotes: 1

Views: 137

Answers (1)

Eran
Eran

Reputation: 394156

Since s0.toCharArray() is a primitive array (char[]), Arrays.asList(s0.toCharArray()) is a List whose single element is a char[]. Similarly, new HashSet<>(Arrays.asList(s1.toCharArray())) is a Set that contains a single char[].

Those two char[] objects are not equal, since arrays don't override Object's equals and each call to toCharArray[] creates a new array isntance, so even if s0==s1, s0.toCharArray() != s1.toCharArray() and s0.toCharArray().equals(s1.toCharArray()) is false.

In your first line of code, I'm assuming that ArrayUtils.toObject converts the primitive array to a reference array (i.e. from char[] to Character[]), which is why containsAll returns true (since both the HashSet and the List contain 4 Character instances, and the Character class overrides equals, so all the Characters in the List are contained in the HashSet).

Upvotes: 3

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