Java: getting a object from a Hashmap by key

Question

I try to refresh my Java skills. But now I stuck at a hashmap. I don't get the right object back. Here is my example code:

public class Sample {

    private static Map<String, Map<String, String>> peaks = new HashMap<String, Map<String, String>>();
    private final String name;

    public Sample(String name) {
        this.name = name;
        this.peaks = new HashMap<String, Map<String, String>>();
    }

    public static Map<String, Map<String, String>> getPeaks() {
        return peaks;}

    public static void addPeak(String peakName, String value) {
        Map<String, String> peak = new HashMap<String, String>();
        peak.put("value", value);
        peaks.put(peakName, peak);
    }
}

public class Main {
    static Map<Integer, Sample> sample = new HashMap<Integer, Sample>();

    public static void main(String[] args) {
        Sample sam = new Sample("Test1");
        sample.put(1, sam);
        sample.get(1).addPeak("A", "1");

        sam = new Sample("Test2");
        sample.put(2, sam);
        sample.get(2).addPeak("B", "123");

        System.out.println(sample.get(1).getPeaks().toString());    
        System.out.println(sample.get(2).getPeaks().toString());    
        System.out.println(sample.get(4).getPeaks().toString());    
    }
}

EVERYTIME THE OUTPUT IS: {B={value=123}}

I don't know which part is wrong. Did I miss something?

@freak It's there, buried. Joko, please post the output as a separate block instead of hiding it inside a comment. — chrylis -cautiouslyoptimistic-
– chrylis -cautiouslyoptimistic-, Commented Sep 25, 2013 at 12:14
If I only remove static I get much more errors (but this is another topic then). I will read about static... — Joko
– Joko, Commented Sep 25, 2013 at 12:20

Arnaud Denoyelle · Accepted Answer · 2013-09-25 12:21:27Z

1

This map is static :

private static Map<String, Map<String, String>> peaks = new HashMap<String, Map<String, String>>();

Hence, every call to Sample.getPeaks() points to the same Map.

//getPeaks() returns the same map in both cases.
// In other words : sample.get(1).getPeaks() == sample.get(2).getPeaks()
System.out.println(sample.get(1).getPeaks().toString());    
System.out.println(sample.get(2).getPeaks().toString());

// This is equivalent to
Map<String, Map<String, String>> myStaticMap = Sample.getPeaks();
System.out.println(myStaticMap.toString());    
System.out.println(myStaticMap.toString());

answered Sep 25, 2013 at 12:21

Arnaud Denoyelle

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Comments

isnot2bad · Accepted Answer · 2013-09-25 12:21:48Z

1

The methods getPeaks and addPeaks as well as the map peaks are static! That means they only exist once (per class), not once per instance of Sample.

So everytime you create a new Sample instance, you're overwriting the static field peaks in the constructor! Remove the static keyword!

answered Sep 25, 2013 at 12:21

isnot2bad

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Comments

Meesh · Accepted Answer · 2013-09-25 17:35:27Z

0

Perhaps you were trying to learn how to use nested Maps, but for this implementation, couldn't you simply create a Map<String, String>? The reason I'm asking is because in addPeak you've made a call to peak.put("value", value), with the hardcoded string "value", which may be unnecessary. Here's some code which may satisfy what you're interested in:

public class WhyMapOfMap {

    // Simpler map
    private Map<String, String> peaks;
    private String name;

    public WhyMapOfMap(String name) {
        this.name = name;
        peaks = new HashMap<>();
    }

    public Map<String, String> getPeaks() {
        return peaks;
    }

    public void addPeak(String peakName, String value) {
        peaks.put(peakName, value);
    }
}

Then you may access the map normally sample.get(key).getPeaks().

answered Sep 25, 2013 at 17:35

Meesh

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2 Comments

Joko Over a year ago

Thank you for your answer! Maybe there is an easier way to do so. But in my complete sourcecode I save more the than only 'value'.

Meesh Over a year ago

Ah I understand. I sometimes find it difficult to model the data I need into standard collections. In those cases I'll favor an approach of object composition. A guiding principles I look to is "smart data and dumb algorithms". Good luck!

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