给出2个Class对象,如何获取Map的Class对象? 例如,假设我有:
Class keyClass = Long.class;
Class valueClass = String.class;
如何获取Map<Long,String>
的 Class 对象?
答案 0 :(得分:0)
没有Map<Long, String>
这样的类。您想要的是Map.class
。 (或HashMap.class
等)
Map<String, Integer> map1 = new HashMap<>();
Map<Long, String> map2 = new HashMap<>();
System.out.println(map1.getClass().equals(map2.getClass()));
结果为true
。
答案 1 :(得分:0)
Map<Long, String>
不是类,但它是一种类型,确切地说,ParameterizedType
是可悲的,用于构造它们的Java代码是私有的,但是它们是获得它的2种方法,更动态的方法是实现该接口:
final class ParameterizedTypeImpl implements ParameterizedType {
private final Type[] actualTypeArguments;
private final Class rawType;
@Nullable private final Type ownerType;
ParameterizedTypeImpl(Class rawType, Type[] actualTypeArguments, @Nullable Type ownerType) {
this.actualTypeArguments = actualTypeArguments.clone();
this.rawType = rawType;
if ((ownerType != null) || (rawType.getDeclaringClass() == null)) {
this.ownerType = ownerType;
}
else {
Class declaringClass = rawType.getDeclaringClass();
if (Modifier.isStatic(rawType.getModifiers())) {
this.ownerType = declaringClass;
}
else {
TypeVariable[] typeParameters = declaringClass.getTypeParameters();
if (typeParameters.length == 0) {
this.ownerType = declaringClass;
}
else {
this.ownerType = new ParameterizedTypeImpl(declaringClass, typeParameters, null);
}
}
}
}
@Override
public Type[] getActualTypeArguments() { return this.actualTypeArguments.clone(); }
@Override
public Class getRawType() { return this.rawType; }
@Nullable @Override
public Type getOwnerType() { return this.ownerType; }
@Override public boolean equals(Object o) {
if (o instanceof ParameterizedType) {
ParameterizedType that = (ParameterizedType) o;
if (this == that) return true;
Type thatOwner = that.getOwnerType();
Type thatRawType = that.getRawType();
return Objects.equals(this.ownerType, thatOwner) && Objects.equals(this.rawType, thatRawType) && Arrays.equals(this.actualTypeArguments, that.getActualTypeArguments());
}
return false;
}
@Override
public int hashCode() {
return Arrays.hashCode(this.actualTypeArguments) ^ Objects.hashCode(this.ownerType) ^ Objects.hashCode(this.rawType);
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder(256);
if (this.ownerType != null) {
sb.append(this.ownerType.getTypeName());
sb.append("$");
if (this.ownerType instanceof ParameterizedTypeImpl) {
sb.append(this.rawType.getName().replace(((ParameterizedTypeImpl) this.ownerType).rawType.getName() + "$", ""));
}
else {
sb.append(this.rawType.getSimpleName());
}
}
else {
sb.append(this.rawType.getName());
}
StringJoiner joiner = new StringJoiner(", ", "<", ">");
joiner.setEmptyValue("");
for (Type type : this.actualTypeArguments) {
joiner.add(type.getTypeName());
}
sb.append(joiner.toString());
return sb.toString();
}
}
然后您可以new ParameterizedTypeImpl(Map.class, new Type[]{String.class, Long.class}, null)
进行操作,注意使该类对其他人不可见,并创建一些工厂方法是一个好习惯。
其他不太动态的方法是使用类型标记,例如gson:
public class TypeToken<T> {
final Type type;
protected TypeToken() {
this.type = this.getClass().getGenericSuperclass();
}
public final Type getType() { return this.type; }
@Override public final int hashCode() { return this.type.hashCode(); }
@Override public final boolean equals(Object o) { return (o instanceof TypeToken<?>) && this.type.equals(((TypeToken<?>) o).type); }
@Override public final String toString() { return this.type.toString(); }
}
,然后是new TypeToken<Map<String, Long>>{}.getType();
-需要在编译时提供类型。
应该有一些提供这两种方法的库,但是我现在还不知道,因为我需要自己编写自己的库以支持从字符串解析。