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Is it bad practice to use mutable objects as Hashmap keys? What happens when you try to retrieve a value from a Hashmap using a key that has been modified enough to change its hashcode?
For example, given
class Key
{
int a; //mutable field
int b; //mutable field
public int hashcode()
return foo(a, b);
// setters setA and setB omitted for brevity
}
with code
HashMap<Key, Value> map = new HashMap<Key, Value>();
Key key1 = new Key(0, 0);
map.put(key1, value1); // value1 is an instance of Value
key1.setA(5);
key1.setB(10);
What happens if we now call map.get(key1)? Is this safe or advisable? Or is the behavior dependent on the language?
It has been noted by many well respected developers such as Brian Goetz and Josh Bloch that :
If an object’s hashCode() value can change based on its state, then we
must be careful when using such objects as keys in hash-based
collections to ensure that we don’t allow their state to change when
they are being used as hash keys. All hash-based collections assume
that an object’s hash value does not change while it is in use as a
key in the collection. If a key’s hash code were to change while it
was in a collection, some unpredictable and confusing consequences
could follow. This is usually not a problem in practice — it is not
common practice to use a mutable object like a List as a key in a
HashMap.
This is not safe or advisable. The value mapped to by key1 can never be retrieved. When doing a retrieval, most hash maps will do something like
Object get(Object key) {
int hash = key.hashCode();
//simplified, ignores hash collisions,
Entry entry = getEntry(hash);
if(entry != null && entry.getKey().equals(key)) {
return entry.getValue();
}
return null;
}
In this example, key1.hashcode() now points to the wrong bucket of the hash table, and you will not be able to retrieve value1 with key1.
If you had done something like,
Key key1 = new Key(0, 0);
map.put(key1, value1);
key1.setA(5);
Key key2 = new Key(0, 0);
map.get(key2);
This will also not retrieve value1, as key1 and key2 are no longer equal, so this check
if(entry != null && entry.getKey().equals(key))
will fail.
Hash maps use hash code and equality comparisons to identify a certain key-value pair with a given key. If the has map keeps the key as a reference to the mutable object, it would work in the cases where the same instance is used to retrieve the value. Consider however, the following case:
T keyOne = ...;
T keyTwo = ...;
// At this point keyOne and keyTwo are different instances and
// keyOne.equals(keyTwo) is true.
HashMap myMap = new HashMap();
myMap.push(keyOne, "Hello");
String s1 = (String) myMap.get(keyOne); // s1 is "Hello"
String s2 = (String) myMap.get(keyTwo); // s2 is "Hello"
// because keyOne equals keyTwo
mutate(keyOne);
s1 = myMap.get(keyOne); // returns "Hello"
s2 = myMap.get(keyTwo); // not found
The above is true if the key is stored as a reference. In Java usually this is the case. In .NET for instance, if the key is a value type (always passed by value), the result will be different:
T keyOne = ...;
T keyTwo = ...;
// At this point keyOne and keyTwo are different instances
// and keyOne.equals(keyTwo) is true.
Dictionary myMap = new Dictionary();
myMap.Add(keyOne, "Hello");
String s1 = (String) myMap[keyOne]; // s1 is "Hello"
String s2 = (String) myMap[keyTwo]; // s2 is "Hello"
// because keyOne equals keyTwo
mutate(keyOne);
s1 = myMap[keyOne]; // not found
s2 = myMap[keyTwo]; // returns "Hello"
Other technologies might have other different behaviors. However, almost all of them would come to a situation where the result of using mutable keys is not deterministic, which is very very bad situation in an application - a hard to debug and even harder to understand.
If key’s hash code changes after the key-value pair (Entry) is stored in HashMap, the map will not be able to retrieve the Entry.
Key’s hashcode can change if the key object is mutable. Mutable keys in HahsMap can result in data loss.
This will not work. You are changing the key value, so you are basically throwing it away. Its like creating a real life key and lock, and then changing the key and trying to put it back in the lock.
As others explained, it is dangerous.
A way to avoid that is to have a const field giving explicitly the hash in your mutable objects (so you would hash on their "identity", not their "state"). You might even initialize that hash field more or less randomly.
Another trick would be to use the address, e.g. (intptr_t) reinterpret_cast<void*>(this) as a basis for hash.
In all cases, you have to give up hashing the changing state of the object.
There are two very different issues that can arise with a mutable key depending on your expectation of behavior.
First Problem: (probably most trivial--but hell it gave me problems that I didn't think about!)
You are attempting to place key-value pairs into a map by updating and modifying the same key object. You might do something like Map<Integer, String> and simply say:
int key = 0;
loop {
map.put(key++, newString);
}
I'm reusing the "object" key to create a map. This works fine in Java because of autoboxing where each new value of key gets autoboxed to a new Integer object. What would not work is if I created my own (mutable) Integer object:
MyInteger {
int value;
plusOne(){
value++;
}
}
Then tried the same approach:
MyInteger key = new MyInteger(0);
loop{
map.put(key.plusOne(), newString)
}
My expectation is that, for instance, I map 0 -> "a" and 1 -> "b". In the first example, if I change int key = 0, the map will (correctly) give me "a". For simplicity let's assume MyInteger just always returns the same hashCode() (if you can somehow manage to create unique hashCode values for all possible states of an object, this will not be an issue, and you deserve an award). In this case, I call 0 -> "a", so now the map holds my key and maps it to "a", I then modify key = 1 and try to put 1 -> "b". We have a problem! The hashCode() is the same, and the only key in the HashMap is my MyInteger key object which has just been modified to be equal to 1, so It overwrites that key's value so that now, instead of a map with 0 -> "a" and 1 -> "b", I have 1 -> "b" only! Even worse, if I change back to key = 0, the hashCode points to 1 -> "b", but since the HashMap's only key is my key object, it satisfied the equality check and returns "b", not "a" as expected.
If, like me, you fall prey to this type of issue, it's incredibly difficult to diagnose. Why? Because if you have a decent hashCode() function it will generate (mostly) unique values. The hash value will largely take care of the inequality problem when structuring the map but if you have enough values, eventually you'll get a collision on the hash value and then you get unexpected and largely inexplicable results. The resultant behavior is that it works for small runs but fails for larger ones.
Advice:
To find this type of issue, modify the hashCode() method, even trivially (i.e. = 0--obviously when doing this, keep in mind that the hash values should be the same for two equal objects*), and see if you get the same results--because you should and if you don't, there's likely a semantic error with your implementation that's using a hash table.
*There should be no danger (if there is--you have a semantic problem) in always returning 0 from a hashCode() (although it would defeat the purpose of a Hash Table). But that's sort of the point: the hashCode is a "quick and easy" equality measure that's not exact. So two very different objects could have the same hashCode() yet not be equal. On the other hand, two equal objects must always have the same hashCode() value.
p.s. In Java, from my understanding, if you do such a terrible thing (as have many hashCode() collisions), it will start using a red-black-tree as opposed to ArrayList. So when you expect O(1) lookup, you'll get O(log(n))--which is better than the ArrayList which would give O(n).
Second Problem:
This is the one that most others seem to be focusing on, so I'll try to be brief. In this use case, I try to map a key-value pair and then I do some work on the key and then want to come back and get my value.
Expectation: key -> value is mapped, I then modify key and try to get(key). I expect that will give me value.
It seems kind of obvious to me that this wouldn't work but I'm not above having tried to use things like Collections as a key before (and quite quickly realizing it doesn't work). It doesn't work because it's quite likely that the hash value of key has changed so you won't even be looking in the correct bucket.
This is why it's very inadvisable to use collections as keys. I would assume, if you were doing this, you're trying to establish a many-to-one relationship. So I have a class (as in teaching) and I want two groups to do two different projects. What I want is that given a group, what is their project? Simple, I divide the class in two, and I have group1 -> project1 and group2 -> project2. But wait! A new student arrives so I place them in group1. The problem is that group1 has now been modified and likely its hash value has changed, therefore trying to do get(group1) is likely to fail because it will look in a wrong or non-existent bucket of the HashMap.
The obvious solution to the above is to chain things--instead of using the groups as keys, give them labels (that don't change) that point to the group and therefore the project: g1 -> group1 and g1 -> project1, etc.
p.s.
Please make sure to define a hashCode() and equals(...) method for any object you expect to use as a key (eclipse and, I'm assuming, most IDE's can do this for you).
Code Example:
Here is a class which exhibits the two different "problem" behaviors. In this case, I attempt to map 0 -> "a", 1 -> "b", and 2 -> "c" (in each case). In the first problem, I do that by modifying the same object, in the second problem, I use unique objects, and in the second problem "fixed" I clone those unique objects. After that I take one of the "unique" keys (k0) and modify it to attempt to access the map. I expect this will give me a, b, c and null when the key is 3.
However, what happens is the following:
map.get(0) map1: 0 -> null, map2: 0 -> a, map3: 0 -> a
map.get(1) map1: 1 -> null, map2: 1 -> b, map3: 1 -> b
map.get(2) map1: 2 -> c, map2: 2 -> a, map3: 2 -> c
map.get(3) map1: 3 -> null, map2: 3 -> null, map3: 3 -> null
The first map ("first problem") fails because it only holds a single key, which was last updated and placed to equal 2, hence why it correctly returns "c" when k0 = 2 but returns null for the other two (the single key doesn't equal 0 or 1). The second map fails twice: the most obvious is that it returns "b" when I asked for k0 (because it's been modified--that's the "second problem" which seems kind of obvious when you do something like this). It fails a second time when it returns "a" after modifying k0 = 2 (which I would expect to be "c"). This is more due to the "first problem": there's a hash code collision and the tiebreaker is an equality check--but the map holds k0, which it (apparently for me--could theoretically be different for someone else) checked first and thus returned the first value, "a" even though had it kept checking, "c" would have also been a match. Finally, the 3rd map works perfectly because I'm enforcing that the map holds unique keys no matter what else I do (by cloning the object during insertion).
I want to make clear that I agree, cloning is not a solution! I simply added that as an example of why a map needs unique keys and how enforcing unique keys "fixes" the issue.
public class HashMapProblems {
private int value = 0;
public HashMapProblems() {
this(0);
}
public HashMapProblems(final int value) {
super();
this.value = value;
}
public void setValue(final int i) {
this.value = i;
}
#Override
public int hashCode() {
return value % 2;
}
#Override
public boolean equals(final Object o) {
return o instanceof HashMapProblems
&& value == ((HashMapProblems) o).value;
}
#Override
public Object clone() {
return new HashMapProblems(value);
}
public void reset() {
this.value = 0;
}
public static void main(String[] args) {
final HashMapProblems k0 = new HashMapProblems(0);
final HashMapProblems k1 = new HashMapProblems(1);
final HashMapProblems k2 = new HashMapProblems(2);
final HashMapProblems k = new HashMapProblems();
final HashMap<HashMapProblems, String> map1 = firstProblem(k);
final HashMap<HashMapProblems, String> map2 = secondProblem(k0, k1, k2);
final HashMap<HashMapProblems, String> map3 = secondProblemFixed(k0, k1, k2);
for (int i = 0; i < 4; ++i) {
k0.setValue(i);
System.out.printf(
"map.get(%d) map1: %d -> %s, map2: %d -> %s, map3: %d -> %s",
i, i, map1.get(k0), i, map2.get(k0), i, map3.get(k0));
System.out.println();
}
}
private static HashMap<HashMapProblems, String> firstProblem(
final HashMapProblems start) {
start.reset();
final HashMap<HashMapProblems, String> map = new HashMap<>();
map.put(start, "a");
start.setValue(1);
map.put(start, "b");
start.setValue(2);
map.put(start, "c");
return map;
}
private static HashMap<HashMapProblems, String> secondProblem(
final HashMapProblems... keys) {
final HashMap<HashMapProblems, String> map = new HashMap<>();
IntStream.range(0, keys.length).forEach(
index -> map.put(keys[index], "" + (char) ('a' + index)));
return map;
}
private static HashMap<HashMapProblems, String> secondProblemFixed(
final HashMapProblems... keys) {
final HashMap<HashMapProblems, String> map = new HashMap<>();
IntStream.range(0, keys.length)
.forEach(index -> map.put((HashMapProblems) keys[index].clone(),
"" + (char) ('a' + index)));
return map;
}
}
Some Notes:
In the above it should be noted that map1 only holds two values because of the way I set up the hashCode() function to split odds and evens. k = 0 and k = 2 therefore have the same hashCode of 0. So when I modify k = 2 and attempt to k -> "c" the mapping k -> "a" gets overwritten--k -> "b" is still there because it exists in a different bucket.
Also there are a lot of different ways to examine the maps in the above code and I would encourage people that are curious to do things like print out the values of the map and then the key to value mappings (you may be surprised by the results you get). Do things like play with changing the different "unique" keys (i.e. k0, k1, and k2), try changing the single key k. You could also see how even the secondProblemFixed isn't actually fixed because you could also gain access to the keys (for instance via Map::keySet) and modify them.
I won't repeat what others have said. Yes, it's inadvisable. But in my opinion, it's not overly obvious where the documentation states this.
You can find it on the JavaDoc for the Map interface:
Note: great care must be exercised if mutable objects are used as map
keys. The behavior of a map is not specified if the value of an object
is changed in a manner that affects equals comparisons while the
object is a key in the map
Behaviour of a Map is not specified if value of an object is changed in a manner that affects equals comparision while object(Mutable) is a key. Even for Set also using mutable object as key is not a good idea.
Lets see a example here :
public class MapKeyShouldntBeMutable {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Map<Employee,Integer> map=new HashMap<Employee,Integer>();
Employee e=new Employee();
Employee e1=new Employee();
Employee e2=new Employee();
Employee e3=new Employee();
Employee e4=new Employee();
e.setName("one");
e1.setName("one");
e2.setName("three");
e3.setName("four");
e4.setName("five");
map.put(e, 24);
map.put(e1, 25);
map.put(e2, 26);
map.put(e3, 27);
map.put(e4, 28);
e2.setName("one");
System.out.println(" is e equals e1 "+e.equals(e1));
System.out.println(map);
for(Employee s:map.keySet())
{
System.out.println("key : "+s.getName()+":value : "+map.get(s));
}
}
}
class Employee{
String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#Override
public boolean equals(Object o){
Employee e=(Employee)o;
if(this.name.equalsIgnoreCase(e.getName()))
{
return true;
}
return false;
}
public int hashCode() {
int sum=0;
if(this.name!=null)
{
for(int i=0;i<this.name.toCharArray().length;i++)
{
sum=sum+(int)this.name.toCharArray()[i];
}
/*System.out.println("name :"+this.name+" code : "+sum);*/
}
return sum;
}
}
Here we are trying to add mutable object "Employee" to a map. It will work good if all keys added are distinct.Here I have overridden equals and hashcode for employee class.
See first I have added "e" and then "e1". For both of them equals() will be true and hashcode will be same. So map sees as if the same key is getting added so it should replace the old value with e1's value. Then we have added e2,e3,e4 we are fine as of now.
But when we are changing the value of an already added key i.e "e2" as one ,it becomes a key similar to one added earlier. Now the map will behave wired. Ideally e2 should replace the existing same key i.e e1.But now map takes this as well. And you will get this in o/p :
is e equals e1 true
{Employee#1aa=28, Employee#1bc=27, Employee#142=25, Employee#142=26}
key : five:value : 28
key : four:value : 27
key : one:value : 25
key : one:value : 25
See here both keys having one showing same value also. So its unexpected.Now run the same programme again by changing e2.setName("diffnt"); which is e2.setName("one"); here ...Now the o/p will be this :
is e equals e1 true
{Employee#1aa=28, Employee#1bc=27, Employee#142=25, Employee#27b=26}
key : five:value : 28
key : four:value : 27
key : one:value : 25
key : diffnt:value : null
So by adding changing the mutable key in a map is not encouraged.
To make the answer compact:
The root cause is that HashMap calculates an internal hash of the user's key object hashcode only once and stores it inside for own needs.
All other operations for data navigation inside the map are doing by this pre-calculated internal hash.
So if you change the hashcode of the key object (mutate) it will be still stored nicely inside the map with the changed key object's hashcode (you could even observe it via HashMap.keySet() and see the altered hashcode).
But HashMap internal hash will not be recalculated of course and it will be the old stored one and the map won't be able to locate your data by the provided mutated key object new hashcode. (e.g. by HashMap.get() or HashMap.containsKey()).
Your key-value pairs will be still inside the map but to get it back you will need that old hash code value that was given when you put your data into the map.
Notice that you also will be unable to get data back by the mutated key object taken right from the HashMap.keySet().
I have a HashMap with key of type Double and my custom object as value.
It looks like this:
private static Map<Double, Incident> incidentHash = new HashMap<>();
The Incident object has following attributes: String date, String address, String incidentType.
Now I have a String date that I get from the user as input and I want to check if there exists any incident in the HashMap with that user inputted date. There can be many Incidents in the HashMap with the given date but as long as there's at least one Incident with the given date, I can do *
something.
I can just iterate over all the values in the HashMap and check if a given date exists but I was wondering if there is any better and more efficient way possible without modifying the data structure.
Given your HashMap, NO, there is not another way of doing so without iterating that HashMap.
As for changing the structure, you could do as Map<String, List<Incident>> that way you would have a date as key and a List of incidents for that date, given your requirement: There can be many Incidents in the HashMap with the given date.
So this would be a O(1)
//considering that the key is added when you have at least one incident
if (yourHash.get("yourDateStringWhatEverTheFormatIs") != null)
You can use streams API (from Java8) as shown in the below code with inline comments:
String userInput="10-APR-2017";
Optional<Map.Entry<Double, Incident>> matchedEntry =
incidentHash.entrySet().stream().
//filter with the condition to match
filter(element -> element.getValue().getDate().equals(userInput)).findAny();
//if the entry is found, do your logic
matchedEntry.ifPresent(value -> {
//do something here
});
If you are looking for something prior to JDK1.8, you can refer the below code:
String userInput="10-APR-2017";
Set<Map.Entry<Double, Incident>> entries = incidentHash.entrySet();
Map.Entry<Double, Incident> matchedEntry = null;
for(Iterator<Map.Entry<Double, Incident>> iterator = entries.iterator();
iterator.hasNext();) {
Map.Entry<Double, Incident> temp = iterator.next();
if(temp.getValue().getDate().equals(userInput)) {
matchedEntry = temp;
break;
}
}
You can use a TreeMap with your custom Comparator. In your Comparator compare the values of dates.
You would have to iterate through the map until you find a data that matches. Since you only need to know if any occurrences exist you can simply exit the loop when you find a match instead of iterating the rest of the map.
You can only keep a second Hash/TreeMap that matches the attribute to the object, so you can also check this attibute qickly. But you have to curate one such map for each attribute you want to access quickly. This makes it a bit more complex and use more memory, but can be much much faster.
If this is not an option the stream API referenced in other answers is a nice and tidy way to iterate over all objects to search for an attribute.
private static Map<Double, Incident> incidentHash = new HashMap<>();
private static Map<String, List<Incident>> incidentsPerDayMap = new HashMap<>();
Given that you don't want to iterate the Map and currently it's the only way to get the required value, I would recommend recomment another Map that contains Date as key and List<Incident> as value. It can be a TreeMap, e.g.:
Map<Date, List<Incident>> incidents = new TreeMap<>();
You can put the entry in this Map whenever an entry is added into the original Map, e.g.:
Incident incident = ;// incident object
Date date; //Date
incidents.computeIfAbsent(date, t -> new ArrayList<>()).add(incident);
Once the user inputs the Date, you can get all the incidents belonging to this date just by incidents.get(). Although that will give you a list and you still need to iterate over it, it will contain a lot less elements and get method in TreeMap will guarantee you log n complexity as it is sorted. So, your search operation will be much more efficient.
I'm checking to see if a key in my HashMap exists, if it does, I also want to check to see if any other keys have a value with the same name as that of the original key I checked for or not.
For example I have this.
System.out.println("What course do you want to search?");
String searchcourse = input.nextLine();
boolean coursefound = false;
if(hashmap.containsKey(searchcourse) == true){
coursefound = true;
}
This checks to see if the key exists in my hashmap, but now I need to check every single key's values for a specific value, in this case the string searchcourse.
Usually I would use a basic for loop to iterate through something like this, but it doesn't work with HashMaps. My values are also stored in a String ArrayList, if that helps.
You will want to look at each entry in the HashMap. This loop should check the contents of the ArrayList for your searchcourse and print out the key that contained the value.
for (Map.Entry<String,ArrayList> entries : hashmap.entrySet()) {
if (entries.getValue().contains(searchcourse)) {
System.out.println(entries.getKey() + " contains " + searchcourse);
}
}
Here are the relevant javadocs:
Map.Entry
HashMap entrySet method
ArrayList contains method
You can have a bi-directional map. E.g. you can have a Map<Value, Set<Key>> or MultiMap for the values to keys or you can use a bi-directional map which is planned to be added to Guava.
As I understand your question, the values in your Map are List<String>. That is, your Map is declares as Map<String, List<String>>. If so:
for (List<String> listOfStrings : myMap.values()) [
if (listOfStrings .contains(searchcourse) {
// do something
}
}
If the values are just Strings, i.e. the Map is a Map<String, String>, then #Matt has the simple answer.
I want to write a program that prints out entries "0" and "4" of the HashMap (i.e. entry.getKey(0) and entry.getKey(4) but it won't let me do this) What would be another way using what I already have?
Basically I have this:
HashMap<String, Integer> hm = new HashMap<String, Integer>();
I can iterate over each entry using this code:
for (Map.Entry<String,Integer> entry : hm.entrySet())
{
System.out.println(entry.getKey() + "/" + entry.getValue());
}
Since people have asked for more contextual information, I am storing a set of strings in the HashMap. For example the 0th entry is "Bob", the 1st entry is "Mindy", the 2nd is "Yasser", the 3rd is "Greg" and the 4th is "Jacky." I want the program to print out the 0th and 4th entries of the populated HashMap.
If you are specific about the keys in the Map, you can directly use get() method.Like this,
Integer value = hm.get("0");
If you want to iterate then use something like the code below :
HashMap<String, Integer> hm = new HashMap<String, Integer>();
for (Entry<String, Integer> entry : hm.entrySet())
{
String key = entry.getKey();
if(key.equals("0") || key.equals("1"))
System.out.println(key + "/" + entry.getValue());
}
You cannot pass index to the getKey() method like getKey(0) etc. Refer the documentation.
HashMap class makes no guarantees as to the order of the map; in particular, it does not guarantee that the order will remain constant over time. So, if you are looking to fetch values from a HashMap based on index, probably it is not possible. Closest to your requirement will be something like LinkedHashMap, which maintains the order of keys for insertion/access order.
HashMap works on principle of hashing, we have put(key, value) and get(key) method for storing and retrieving Objects from HashMap. When we pass Key and Value object to put() method on Java HashMap, HashMap implementation calls hashCode() method on Key object and applies returned hashcode() into its own hashing function to find a bucket location for storing Entry object, important point to mention is that HashMap in Java stores both key and value object as Map.Entry in bucket which is essential to understand the retrieving logic.
The HashMap has no defined ordering of keys. You may use LinkedHashMap instead of HashMap It will always return keys in same order (as insertion) when calling keySet(). And then you pick the 0th or 4th key.Later you can retrieve the value for the keys you fetched at 0th and 4th location.
I would recommend simply using the get() method with the provided key. Iteration is not necessary in this case.
Since people have asked for more contextual information, I am storing a set of strings in the HashMap. For example the 0th entry is "Bob", the 1st entry is "Mindy", the 2nd is "Yasser", the 3rd is "Greg" and the 4th is "Jacky." I want the program to print out the 0th and 4th entries of the populated HashMap.
With how it is being used, an HashMap makes no sense. Use a String[] instead!
String[] names = {"Bob", "Mindy", "Yasser", "Greg", "Jacky" };
System.out.println("When " + names[0] + " met " + names[4]);
Why not a targeted approach:
String[] targets = {"0", "4"};
for (String target : targets) {
System.out.println(target + "/" + hm.get(target));
}
Way more efficient than the "big hammer" full iteration approach, and you get output order for free.
If you have a lot of entries in your Map you may considerate to use one of the navigable collections, like TreeMap.
NavigableMap<String,String> map = new TreeMap<>()
map.subMap("0", true, "4", true);
Visit: http://docs.oracle.com/javase/6/docs/api/java/util/NavigableMap.html
If I pass the same key multiple times to HashMap’s put method, what happens to the original value? And what if even the value repeats? I didn’t find any documentation on this.
Case 1: Overwritten values for a key
Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","not one");
mymap.put("1","surely not one");
System.out.println(mymap.get("1"));
We get surely not one.
Case 2: Duplicate value
Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","not one");
mymap.put("1","surely not one");
// The following line was added:
mymap.put("1","one");
System.out.println(mymap.get("1"));
We get one.
But what happens to the other values? I was teaching basics to a student and I was asked this. Is the Map like a bucket where the last value is referenced (but in memory)?
By definition, the put command replaces the previous value associated with the given key in the map (conceptually like an array indexing operation for primitive types).
The map simply drops its reference to the value. If nothing else holds a reference to the object, that object becomes eligible for garbage collection. Additionally, Java returns any previous value associated with the given key (or null if none present), so you can determine what was there and maintain a reference if necessary.
More information here: HashMap Doc
You may find your answer in the javadoc of Map#put(K, V) (which actually returns something):
public V put(K key,
V value)
Associates the specified value with the specified key in this map
(optional operation). If the map
previously contained a mapping for
this key, the old value is replaced by
the specified value. (A map m is said
to contain a mapping for a key k if
and only if m.containsKey(k) would
return true.)
Parameters:
key - key with which the specified value is to be associated.
value - value to be associated with the specified key.
Returns:
previous value associated with specified key, or null if there was no
mapping for key. (A null return can also indicate that the map previously associated null with the specified key, if the implementation supports null values.)
So if you don't assign the returned value when calling mymap.put("1", "a string"), it just becomes unreferenced and thus eligible for garbage collection.
it's Key/Value feature and you could not to have duplicate key for several values because when you want to get the actual value which one of values is belong to entered keyin your example when you want to get value of "1" which one is it ?!that's reasons to have unique key for every value but you could to have a trick by java standard lib :
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
public class DuplicateMap<K, V> {
private Map<K, ArrayList<V>> m = new HashMap<>();
public void put(K k, V v) {
if (m.containsKey(k)) {
m.get(k).add(v);
} else {
ArrayList<V> arr = new ArrayList<>();
arr.add(v);
m.put(k, arr);
}
}
public ArrayList<V> get(K k) {
return m.get(k);
}
public V get(K k, int index) {
return m.get(k).size()-1 < index ? null : m.get(k).get(index);
}
}
and you could to use it in this way:
public static void main(String[] args) {
DuplicateMap<String,String> dm=new DuplicateMap<>();
dm.put("1", "one");
dm.put("1", "not one");
dm.put("1", "surely not one");
System.out.println(dm.get("1"));
System.out.println(dm.get("1",1));
System.out.println(dm.get("1", 5));
}
and result of prints are :
[one, not one, surely not one]
not one
null
It replaces the existing value in the map for the respective key. And if no key exists with the same name then it creates a key with the value provided.
eg:
Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","two");
OUTPUT
key = "1", value = "two"
So, the previous value gets overwritten.
The prior value for the key is dropped and replaced with the new one.
If you'd like to keep all the values a key is given, you might consider implementing something like this:
import org.apache.commons.collections.MultiHashMap;
import java.util.Set;
import java.util.Map;
import java.util.Iterator;
import java.util.List;
public class MultiMapExample {
public static void main(String[] args) {
MultiHashMap mp=new MultiHashMap();
mp.put("a", 10);
mp.put("a", 11);
mp.put("a", 12);
mp.put("b", 13);
mp.put("c", 14);
mp.put("e", 15);
List list = null;
Set set = mp.entrySet();
Iterator i = set.iterator();
while(i.hasNext()) {
Map.Entry me = (Map.Entry)i.next();
list=(List)mp.get(me.getKey());
for(int j=0;j<list.size();j++)
{
System.out.println(me.getKey()+": value :"+list.get(j));
}
}
}
}
Associates the specified value with the specified key in this map. If the map previously contained a mapping for the key, the old value is replaced.
To your question whether the map was like a bucket: no.
It's like a list with name=value pairs whereas name doesn't need to be a String (it can, though).
To get an element, you pass your key to the get()-method which gives you the assigned object in return.
And a Hashmap means that if you're trying to retrieve your object using the get-method, it won't compare the real object to the one you provided, because it would need to iterate through its list and compare() the key you provided with the current element.
This would be inefficient. Instead, no matter what your object consists of, it calculates a so called hashcode from both objects and compares those. It's easier to compare two ints instead of two entire (possibly deeply complex) objects. You can imagine the hashcode like a summary having a predefined length (int), therefore it's not unique and has collisions. You find the rules for the hashcode in the documentation to which I've inserted the link.
If you want to know more about this, you might wanna take a look at articles on javapractices.com and technofundo.com
regards
Maps from JDK are not meant for storing data under duplicated keys.
At best new value will override the previous ones.
Worse scenario is exception (e.g when you try to collect it as a stream):
No duplicates:
Stream.of("one").collect(Collectors.toMap(x -> x, x -> x))
Ok. You will get: $2 ==> {one=one}
Duplicated stream:
Stream.of("one", "not one", "surely not one").collect(Collectors.toMap(x -> 1, x -> x))
Exception java.lang.IllegalStateException: Duplicate key 1 (attempted merging values one and not one)
| at Collectors.duplicateKeyException (Collectors.java:133)
| at Collectors.lambda$uniqKeysMapAccumulator$1 (Collectors.java:180)
| at ReduceOps$3ReducingSink.accept (ReduceOps.java:169)
| at Spliterators$ArraySpliterator.forEachRemaining (Spliterators.java:948)
| at AbstractPipeline.copyInto (AbstractPipeline.java:484)
| at AbstractPipeline.wrapAndCopyInto (AbstractPipeline.java:474)
| at ReduceOps$ReduceOp.evaluateSequential (ReduceOps.java:913)
| at AbstractPipeline.evaluate (AbstractPipeline.java:234)
| at ReferencePipeline.collect (ReferencePipeline.java:578)
| at (#4:1)
To deal with duplicated keys - use other package, e.g:
https://google.github.io/guava/releases/19.0/api/docs/com/google/common/collect/Multimap.html
There is a lot of other implementations dealing with duplicated keys.
Those are needed for web (e.g. duplicated cookie keys, Http headers can have same fields, ...)
Good luck! :)
I always used:
HashMap<String, ArrayList<String>> hashy = new HashMap<String, ArrayList<String>>();
if I wanted to apply multiple things to one identifying key.
public void MultiHash(){
HashMap<String, ArrayList<String>> hashy = new HashMap<String, ArrayList<String>>();
String key = "Your key";
ArrayList<String> yourarraylist = hashy.get(key);
for(String valuessaved2key : yourarraylist){
System.out.println(valuessaved2key);
}
}
you could always do something like this and create yourself a maze!
public void LOOK_AT_ALL_THESE_HASHMAPS(){
HashMap<String, HashMap<String, HashMap<String, HashMap<String, String>>>> theultimatehashmap = new HashMap <String, HashMap<String, HashMap<String, HashMap<String, String>>>>();
String ballsdeep_into_the_hashmap = theultimatehashmap.get("firststring").get("secondstring").get("thirdstring").get("forthstring");
}
BTW, if you want some semantics such as only put if this key is not exist. you can use concurrentHashMap with putIfAbsent() function.
Check this out:
https://docs.oracle.com/javase/7/docs/api/java/util/concurrent/ConcurrentHashMap.html#put(K,%20V)
concurrentHashMap is thread safe with high performance since it uses "lock striping" mechanism to improve the throughput.
Yes, this means all the 1 keys with value are overwriten with the last added value and here you add "surely not one" so it will display only "surely not one".
Even if you are trying to display with a loop, it will also only display one key and value which have same key.
HashMap<Emp, Emp> empHashMap = new HashMap<Emp, Emp>();
empHashMap.put(new Emp(1), new Emp(1));
empHashMap.put(new Emp(1), new Emp(1));
empHashMap.put(new Emp(1), new Emp());
empHashMap.put(new Emp(1), new Emp());
System.out.println(empHashMap.size());
}
}
class Emp{
public Emp(){
}
public Emp(int id){
this.id = id;
}
public int id;
#Override
public boolean equals(Object obj) {
return this.id == ((Emp)obj).id;
}
#Override
public int hashCode() {
return id;
}
}
OUTPUT : is 1
Means hash map wont allow duplicates, if you have properly overridden equals and hashCode() methods.
HashSet also uses HashMap internally, see the source doc
public class HashSet{
public HashSet() {
map = new HashMap<>();
}
}