Problem using generic map with wildcard

Question

I have a method that returns a map defined as:

public Map<String, ?> getData();

The actual implementation of this method is not clear to me, but, when I try to do:

obj.getData().put("key","value")

I get following compile time error message:

The method put(String, capture#9-of ?) in the type Map is not applicable for the arguments (String, String)

What is the problem? Is String not of type anything?

Thanks in advance.

Solution 1

The wildcard means "the value type parameter could be anything" - it doesn't mean "you can use this as if it were anything you want it to be". In other words, a Map<String, UUID> is valid as a Map<String, ?> - but you wouldn't want to be able to put a String value into it.

If you want a map which can definitely accept string values, you want:

Map<String, ? super String>

Solution 2

The return type of

Map<String, ?>

is the same as

Map<String, ? extends Object>

The means that the concrete type returned could be a Map<String, AnyClass>. You can't put a String into an AnyClass, hence the error.

A good general principle is to not use wildcards in method return types.

Solution 3

Map<String, ?> is a short form of Map<String,? extends Object> and doesn't mean that anything can be added as value. It says that the Map-object can have any generic value type extending Object.

This means that the Map object can be a HashMap<String, String> or a HashMap<String, Integer> as well. Because the compiler can't check which value types will be accepted, he won't let you call methods with the value type as a parameter.

Note:

• You can call methods with the value type as a return value, because everything must extend Object (? extends Object)
• A Map<String, ? super String> will have the opposite effect: You can always use a String as parameter, but the return-type is unclear.

Comments

[Jon Skeet]

A string certainly is an object. What precise difference do you believe it would make to call the String constructor explicitly in this case?

[enguerran]

String is not a primitive type, is it?

[Jon Skeet]

No, it's not a primitive type... so why did you claim it's not an object?

[enguerran]

I believed it was a primitive, like a almost object... Mistaken! But what is the difference between my thought and Ben Lings one?

[Jon Skeet]

Ben Lings' answer is completely different, and accurate. What do you think the similarity is between your answer and Ben's?

[enguerran]

He told us than Map<?> is the same as List<? extends Object>. And he told than AnyClass, as long as AnyClass extends Object, can't be a String. I did some tries: if I write List<?> list = new ArrayList<Object>(), I can't do anything else than list.add(null)... My brain cannot get it...

[enguerran]

OK, it seems I will not have my explanation... I get that List<? extends Object> is a read-only List. I get too that List<? super Object> is the correct answer. But I cannot get why this and that.

[nybon]

String class is final, so just Map<String, String>

[Misa]

@nybon Suppose you have class A extends Map<String, String> and class B extends Map<String, Object>, you want to write a method that can take either class A or class B as argument. That's when you need Map<String, ? super String>