java generics: getting class of a class with generic parameters

Question

I am curious how to make this work

 Class<Map<String,String>> food = Map.class;

That obviously doesn't work. I would want something like this

 Class<Map<String,String>> food = Map<String,String>.class;

but this seems like not a valid java sytax.

How can make this work?

EDIT: The reason I want this is because I have a method like this

   protected <ConfigType> ConfigValue<ConfigType> getSectionConfig(String name, Class<ConfigType> configType) {
        return config.getConfig(name);
    }

I would like to call this as so

ConfigValue<Map<String,Object>> config = getSectionConfig("blah", Map<String,Object>.class>);
Map<String,Value> val = config.value();

Map<String,String>.class can't exist. A Type that describes that signature can though, and you can use a super type token to get that type, but Types don't have the reflection capabilities of a Class. — millimoose
– millimoose, Commented Nov 15, 2013 at 21:23
@JudgeMental Yes but I want to limit casting to just one place. Not all over my code. — Surya
– Surya, Commented Nov 16, 2013 at 17:57

ZhongYu · Accepted Answer · 2013-11-15 22:05:16Z

19

Do a brute cast

    Class<Map<String,String>> clazz = 
             (Class<Map<String,String>>)(Class)Map.class;

this is not theoretically correct, but it is not our fault. We need such hacks some times.

answered Nov 15, 2013 at 22:05

ZhongYu

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rgettman · Accepted Answer · 2013-11-15 21:25:38Z

4

According to the JLS, Section 15.8.2, you can't do that, because:

The type of C.class, where C is the name of a class, interface, or array type (§4.3), is Class<C>.

The closest you can come is

Class<Map> food = Map.class;

answered Nov 15, 2013 at 21:25

rgettman

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1 Comment

Surya Over a year ago

That is what I have currently but its not entirely typesafe and I have to do casting. :-\

Andrei Hryshanovich · Accepted Answer · 2023-09-20 21:28:54Z

0

Not sure but may be ParameterizedTypeReference class can help you?

private static final ParameterizedTypeReference<Map<String, Object>> YOUR_TYPE_NAME = new ParameterizedTypeReference<Map<String, Object>>() {};

answered Sep 20, 2023 at 21:28

Andrei Hryshanovich

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