小编Kev*_*tos的帖子

我试图理解java中的一个程序,它使用lambda将对象返回到接口的引用变量中.我想将lambda转换为简单的java函数,但不知道该怎么做.

该计划如下: -

public class AgentLicenseLazyModel extends CoreDataModel<AgentLicenseModel> {

    public AgentLicenseLazyModel(final List<AgentLicenseModel> entities) {
        super(AgentLicenseModel.class, entities, (sortField, sortOrder) -> {
            return new AgentLicenseLazySorter(sortField, sortOrder);
        });
    }
}

这是抽象类:

public abstract class CoreDataModel<T extends AbstractEntityModel> extends LazyDataModel<T> {

private final Class<T> entityClass;
private final List<T> entities;
private final CoreDataSorterProducer<T> coreDataSorterProducer;

public CoreDataModel(final Class<T> entityClass, final List<T> entities, final CoreDataSorterProducer<T> coreDataSorterProducer) {
    this.entityClass = entityClass;
    this.entities = entities;
    this.coreDataSorterProducer = coreDataSorterProducer;

    if (entities != null) {
        setRowCount(entities.size());
    }
}

如何在不使用lambda进行学习的情况下将此程序转换为简单的java程序.请帮忙.

编辑:这是CoreDataSorterProducer:

@FunctionalInterface …