相关疑难解决方法(0)

看起来很简单,但是,如何将Kotlin初始化MutableList为空MutableList？

我可以通过这种方式破解它,但我确信有更容易的东西:

var pusta: List<Kolory> = emptyList()
var cos: MutableList<Kolory> = pusta.toArrayList()

Abstract控制器类需要REST中的对象列表.使用Spring RestTemplate时,它不会将其映射到所需的类,而是返回Linked HashMAp

 public List<T> restFindAll() {

    RestTemplate restTemplate = RestClient.build().restTemplate();
    ParameterizedTypeReference<List<T>>  parameterizedTypeReference = new ParameterizedTypeReference<List<T>>(){};
    String uri= BASE_URI +"/"+ getPath();

    ResponseEntity<List<T>> exchange = restTemplate.exchange(uri, HttpMethod.GET, null,parameterizedTypeReference);
    List<T> entities = exchange.getBody();
    // here entities are List<LinkedHashMap>
    return entities;

}

如果我用,

ParameterizedTypeReference<List<AttributeInfo>>  parameterizedTypeReference = 
    new ParameterizedTypeReference<List<AttributeInfo>>(){};
    ResponseEntity<List<AttributeInfo>> exchange =
  restTemplate.exchange(uri, HttpMethod.GET, null,parameterizedTypeReference);

它工作正常.但不能放入所有子类,任何其他解决方案.

我想ArgumentCaptor在kotlin.

我所做的：

val c := ArgumentCaptor<List<MyClass>, 
List<MyClass>>.forClass(List<MyClass>::class.java)

但它说

Only classes are allowed on the left hand side of a class literal