How to fix HashMap updating every existing value instead of single value

Question

I've been working on a drunk walker coding problem (custom user classes, etc.), and I'm going insane trying to fix this small issue.

I messed around with the code (to no avail), and so with no hope in sight, I decided to get an outside opinion.

The code I used for adding into the hashmap is like this:

if (hashMap.containsKey(key) == false) {
    hashMap.put(key, 1);
}
else {
    hashMap.put(key, value + 1);
}

In theory, this should be completely fine. If the key is not saved in map, it is added to the map with a value of 1. If the key is actually there in the map, then the value is incremented by one. The key is just an instance of a custom class that takes two integer variables. It is being constantly updated.

At the end of the program, if I display entries in the hashmap with values greater than 1, it should look something like this:

Visited Intersection [avenue=8, street=42] 3 times!
Visited Intersection [avenue=8, street=63] 2 times!

But when I observed what the hashmap looked like with each function call, it looked like this:

Hash Map: {Intersection [avenue=6, street=22]=1}

Hash Map: {Intersection [avenue=6, street=23]=1, Intersection 
[avenue=6, street=23]=1}

Hash Map: {Intersection [avenue=6, street=22]=2, Intersection 
[avenue=6, street=22]=1}

Hash Map: {Intersection [avenue=5, street=22]=2, Intersection 
[avenue=5, street=22]=1, Intersection [avenue=5, street=22]=1}

Hash Map: {Intersection [avenue=6, street=22]=3, Intersection 
[avenue=6, street=22]=1, Intersection [avenue=6, street=22]=1}

...

Every entry in the hashmap was being overwritten, and the end product was this:

Visited Intersection [avenue=8, street=20] 3 times!
Visited Intersection [avenue=8, street=20] 2 times!
Visited Intersection [avenue=8, street=20] 2 times!
Visited Intersection [avenue=8, street=20] 2 times!
...

Originally I thought the code for adding into the hashmap was incorrect, since every key was being overwritten and only the last updated one is displayed, but now I think it has to do with the actual updating of the key.

Penny for your thoughts? Sorry if it's a bit vague.

`hashMap.put(key, value + 1);` <-- where is `value` being defined? — Tim Biegeleisen, Apr 15 '19 at 04:53
I dont understood your problem. What do you want and how look your key-class? And what is your value? — akop, Apr 15 '19 at 04:57
In addition to your actual bug: (1) Don't use `==` with booleans; use `!map.containsKey(key)`. (2) `map.merge(key, value, v -> v + 1)` — chrylis -cautiouslyoptimistic-, Apr 15 '19 at 05:02

Gray · Accepted Answer · 2019-04-15T05:43:49.433

Every entry in the hashmap was being overwritten...

I suspect you don't quite understand how HashMap works. HashMap stores a reference to the key not a copy. I suspect that you are overwriting the fields in the Intersection after you have put it into the map. This is a very bad pattern and can result in some very strange results.

Couple of things to check.

You should be doing a new Intersection(avenue, street) each time.
Consider making the 2 fields in your Intersection be final. This is always a good pattern so you don't change the value of the key unintentionally. For sure if one or both of the fields are "identity" field, it should be final.
You need to make sure that the Intersection object has appropriate hashcode() and equals() methods that correctly identifies each value. Otherwise each Intersection will be stored in the map regardless if they have the same avenue and street values. See my answer here: https://stackoverflow.com/a/9739583/179850
You should get getting the count of the intersections from the map and then incrementing the value.

Maybe something like:

Intersection key = new Intersection(8, 42);
...
Integer count = hashMap.get(key);
if (count == null) {
   hashMap.put(key, 1);
} else {
   hashMap.put(key, value + 1);
}
...
public class Intersection {
   // these fields can't be changed
   private final int avenue;
   private final int street;
   ...

How to fix HashMap updating every existing value instead of single value

1 Answers1