It puts wrong null value from String array in HashMap

Question

i am a newbie in Java (coming from JavaScript developing on Adobe LiveCycle) and facing the following problem:

I have a String array with several items. I want to put only the items with the value "a" to a HashMap. But instead of 3 "a" values in the HashMap i get 1 null value there. Why is that?

    String[] s = {"a", "a", "b", "a"};

    Map m = new HashMap();

    for (int i = 0; i < s.length; i++) {
        if (s[i].equals("a")) {
            m.put(i, s[i]);
        }
    }

    for (int i = 0; i < m.size(); i++) {
        System.out.println(m.get(i));
    }

    // Prints
    //a
    //a
    //null

You have a Map<Integer, String> where key is the index of the string in the array. You shouldn't iterate from 0 to map size, but over map's actual entries. Use for (Integer key : map.keySet()) — Marko Topolnik
– Marko Topolnik, Commented Oct 7, 2016 at 13:38
You put the items with their corresponding index in the Map, i.e. you have a Map with content {0=a, 1=a, 3=a}. Therefore, if you try to access the map with 2 in the m.get(...), you get a null since key 2 is not found in the map. On a sidenote: you are using raw types. You should bind the types of the Map and HashMap properly: Map<Integer, String> m = new HashMap<Integer, String>(); — Turing85
– Turing85, Commented Oct 7, 2016 at 13:39
m.get(i) Returns the value for keyi. there you have no index like an array — Jens
– Jens, Commented Oct 7, 2016 at 13:41

OldCurmudgeon · Accepted Answer · 2016-10-07 13:39:26Z

2

You are putting the items in the map with key 0, 1 and 3.

You are taking them out with key 0, 1, an 2.

Use:

    for (Object o : m.keySet()) {
        System.out.println(m.get(o));
    }

or - better:

    Map<Integer, String> m = new HashMap<>();

    ...

    for (Integer i : m.keySet()) {
        System.out.println(i + " -> " + m.get(i));
    }

answered Oct 7, 2016 at 13:39

OldCurmudgeon

66.1k18 gold badges126 silver badges220 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Turing85 · Accepted Answer · 2016-10-08 07:35:51Z

1

You put the items with their corresponding index in array s in the Map, i.e. you have a Map with content {0=a, 1=a, 3=a}. Therefore if you try to access the map with key 2 (m.get(2)), you get a null since key 2 is not found in m.

Instead of using a for-loop over m's size, I recommend iteration over m's keySet() via a foreach-loop:

for (Object key : m.keySet()) {
    System.out.println("key: " + key + ", value: " + m.get(key));
}

On a sidenote: you are using raw types. You should bind the types of the Map and HashMap properly (see the Javadoc of Map for details): Map<Integer, String> m = new HashMap<Integer, String>();. With properly bound types, key in the for-loop can be of type int or Integer. I recommend type Integer to avoid unnecessary Auto(un)boxing.

edited Oct 8, 2016 at 7:35

answered Oct 7, 2016 at 13:44

Turing85

20.3k7 gold badges41 silver badges66 bronze badges

Comments

Murat Karagöz · Accepted Answer · 2016-10-07 13:47:14Z

1

Your code is working correctly, but you are accessing it not correctly.

String[] s = {"a", "a", "b", "a"};
for (int i = 0; i < s.length; i++) {
    if (s[i].equals("a")) {
        m.put(i, s[i]);
    }
}

This puts it like this

First iteration : m.put(0, "a");
Second iteration : m.put(1, "a");
Third iteration : "b" doest not equal "a" but still counts the index i up
Fourth iteration: m.put(3, "a");

Apart from the other answers you can still use your range based loop and access it with an Iterator

Iterator<String> it = m.values().iterator();
        while (it.hasNext()) {
            System.out.println(it.next());
        }

answered Oct 7, 2016 at 13:47

Murat Karagöz

37.8k16 gold badges82 silver badges114 bronze badges

Collectives™ on Stack Overflow

It puts wrong null value from String array in HashMap

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related