java：不兼容的类型：推理变量 T 具有不兼容的边界等式约束：下限：java.util.List<> 用于添加字段

Question

Hello I want to add a field in HashMap but I have an error

"java: incompatible types: inference variable T has incompatible bounds equality constraints: java.util.Map<java.lang.String,java.lang.Object>lower bounds: java.lang.Object" and I don't see why?

metadataMap.put(review.getField(), metadataMap.get(review.getField()).stream()
                            .filter(suggestion -> review.getTerm() != null && review.getTerm().equals(suggestion.get(Constants.Review.TERM)))
                            .map(termInfo -> termInfo.put(Constants.STATUS, review.getStatus()))
                            .collect(Collectors.toList()));

Here is the other intel about my code

private void removeReviewFromSuggestionsOrDetails(ContainerModel container, ReviewModel review, String field, String instance)
        throws ApiException {
    Map<String, List<Map<String, Object>>> metadataMap = (Map<String, List<Map<String, Object>>>) container.get(field);

Should I use map there or something else? For add my new field Status

Answer 1

Looking at such a small sample of code, I cannot reproduce the problem here and be sure of my solution.

However, I would suggest you to look at your "Types".

At your "removeReviewFromSuggestionsOrDetails" function:

Map<String, List<Map<String, Object>>> metadataMap = 

When declaring this, you are saying:

-this is a Map that receives

-and receives a

Therefore, to proper use the method "put(Key, Values)" you should match the types your Map will recive with the types you use as arguments in the method.

What should be done is:

metadataMap.put(Keys, Values) should be metadataMap.put(String, List<Map<String, Object>>)

What you are doing is:

metadataMap.put(String, Map<String, List<Map<String, Object>>>)

Do you see the problem now? You are giving the put() method a String as Key and a Map as Value.

However, you declared your metadataMap to receive a String as Key and a List of Maps as Value.

I suppose this will fix your problem.