java: HashMap<String, int> not working

JavaGenerics

Java Problem Overview


HashMap<String, int> doesn't seem to work but HashMap<String, Integer> does work. Any ideas why?

Java Solutions


Solution 1 - Java

You can't use primitive types as generic arguments in Java. Use instead:

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

With auto-boxing/unboxing there is little difference in the code. Auto-boxing means you can write:

myMap.put("foo", 3);

instead of:

myMap.put("foo", new Integer(3));

Auto-boxing means the first version is implicitly converted to the second. Auto-unboxing means you can write:

int i = myMap.get("foo");

instead of:

int i = myMap.get("foo").intValue();

The implicit call to intValue() means if the key isn't found it will generate a NullPointerException, for example:

int i = myMap.get("bar"); // NullPointerException

The reason is type erasure. Unlike, say, in C# generic types aren't retained at runtime. They are just "syntactic sugar" for explicit casting to save you doing this:

Integer i = (Integer)myMap.get("foo");

To give you an example, this code is perfectly legal:

Map<String, Integer> myMap = new HashMap<String, Integer>();
Map<Integer, String> map2 = (Map<Integer, String>)myMap;
map2.put(3, "foo");

Solution 2 - Java

GNU Trove support this but not using generics. http://trove4j.sourceforge.net/javadocs/gnu/trove/TObjectIntHashMap.html

Solution 3 - Java

You cannot use primitive types in HashMap. int, or double don't work. You have to use its enclosing type. for an example

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

Now both are objects, so this will work.

Solution 4 - Java

int is a primitive type, you can read what does mean a primitive type in java here, and a Map is an interface that has to objects as input:

public interface Map<K extends Object, V extends Object>

object means a class, and it means also that you can create an other class that exends from it, but you can not create a class that exends from int. So you can not use int variable as an object. I have tow solutions for your problem:

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

or

Map<String, int[]> map = new HashMap<>();
int x = 1;

//put x in map
int[] x_ = new int[]{x};
map.put("x", x_);

//get the value of x
int y = map.get("x")[0];

Solution 5 - Java

You can use reference type in generic arguments, not primitive type. So here you should use

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

and store value as

myMap.put("abc", 5);

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiongklotsView Question on Stackoverflow
Solution 1 - JavacletusView Answer on Stackoverflow
Solution 2 - JavaPeter LawreyView Answer on Stackoverflow
Solution 3 - Javauser6811894View Answer on Stackoverflow
Solution 4 - JavaYounes MeridjiView Answer on Stackoverflow
Solution 5 - JavaChitrapal SinghView Answer on Stackoverflow