[英]How to declare Java List of Object where each one implements Comparable
[英]How to declare Java Comparable in this situation
这是我正在研究的课程:
public class Thing<T extends Comparable<? super T>> {
private Map<String, List<SourcedValue<T>>> properties;
}
然后, SourcedValue
是这样的:
public class SourcedValue<T extends Comparable<? super T>>
implements Comparable<****?*****> {
private T value;
private List<Sources> sources;
@Override
public int compareTo(SourcedValue<****?****> other) {
return value.compareTo(other);
}
}
我怎么把***?***
?
我需要做的是在一些转换方法中对Thing
中的List<SourcedValue<T>>
进行排序,该方法创建Thing
并填充其properties
(以及每个属性的List<SourcedValue<T>>
)。
我不确定这是否会对您有所帮助,但这是我实现您的要求的一种方式。 如果这有助于你,我会很高兴的。
public class SourcedValue<T extends Comparable<? super T>>
implements Comparable<SourcedValue<? super T>> {
private T value;
private List<Integer> sources;
@Override
public int compareTo(SourcedValue<? super T> o) {
return value.compareTo((T)o.value);
}
}
此处使用super
似乎也是多余的。 即使是以下解决方案也应该产生相同的结果
public class SourcedValue<T extends Comparable<T>>
implements Comparable<SourcedValue<T>> {
private T value;
private List<Integer> sources;
@Override
public int compareTo(SourcedValue<T> o) {
return value.compareTo(o.value);
}
}
这是我可能会这样做的:
TypedProperty.java
/**
* A TypedProperty class - more than String.
* @link https://stackoverflow.com/questions/56498727/how-to-declare-java-comparable-in-this-situation
*/
public class TypedProperty<T extends Comparable<T>> implements Comparable<TypedProperty<T>> {
private final T value;
private final Class<T> type;
public TypedProperty(T value, Class<T> clazz) {
this.value = value;
this.type = clazz;
}
public T getValue() {
return value;
}
public Class<T> getType() {
return type;
}
@Override
public int compareTo(TypedProperty<T> other) {
return this.value.compareTo(other.value);
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
TypedProperty<?> that = (TypedProperty<?>) o;
if (!getValue().equals(that.getValue())) return false;
return getType().equals(that.getType());
}
@Override
public int hashCode() {
int result = getValue().hashCode();
result = 31 * result + getType().hashCode();
return result;
}
@Override
public String toString() {
final StringBuffer sb = new StringBuffer("TypedProperty{");
sb.append("value=").append(value);
sb.append(", type=").append(type);
sb.append('}');
return sb.toString();
}
}
一个JUnit测试来证明它的工作原理:
import org.junit.Assert;
import org.junit.Test;
import java.text.DateFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Date;
import java.util.List;
/**
* @link https://stackoverflow.com/questions/56498727/how-to-declare-java-comparable-in-this-situation
*/
public class TypedPropertyTest {
@Test
public void testConstructor_DateTypedProperty() throws ParseException {
// setup
String [] testDates = { "2019-06-07", "2018-03-17", "2017-01-01" };
List<String> expected = Arrays.asList(testDates);
Collections.sort(expected);
DateFormat format = new SimpleDateFormat("yyyy-MM-dd");
List<TypedProperty<Date>> typedProperties = new ArrayList<>();
for (String testDate : testDates) {
typedProperties.add(new TypedProperty<>(format.parse(testDate), Date.class));
}
// test
Collections.sort(typedProperties);
// assert
for (int i = 0; i < typedProperties.size(); ++i) {
Assert.assertEquals(format.parse(expected.get(i)), typedProperties.get(i).getValue());
}
}
}
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