我将如何创建一个存储另一个 class 对象的字段?
How would I go about in creating a field in which store object of another class?
所以这里我有一个 class 代表音乐节表演的表演。
public class Act {
private int num_members;
private String name;
private String kind;
private String stage;
public Act(int num_members, String name, String kind, String stage) {
this.num_members = num_members;
this.name = name;
this.kind = kind;
this.stage = stage;
}
public Act(int num_members, String name, String kind) {
this.num_members = num_members;
this.name = name;
this.kind = kind;
stage = null;
}
public int getNum_members() {
return num_members;
}
public void setNum_members(int num_members) {
this.num_members = num_members;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getKind() {
return kind;
}
public void setKind(String kind) {
this.kind = kind;
}
public String getStage() {
return stage;
}
public void setStage(String stage) {
this.stage = stage;
}
public String toString() {
return "(" + num_members + ", " + name + ", " + kind + ", " + stage + ")";
}
}
在另一个名为 LineUp 的 class 中,我想在一个名为 acts
的字段中存储最多 30 个行为,并且还有一个方法 addAct 将 Act
作为参数并将其添加到 acts
。这是我到目前为止使用的代码,但我不太确定该怎么做。
public class LineUp {
Act[] acts;
public LineUp() {
Act[] acts = new Act[30];
}
void addAct(Act a) {
}
}
如果您使用数组,您必须确保永远不需要添加超过 30 Act
并跟踪数字以帮助您,或者更简单的解决方案是使用 List<Act>
因为你不需要用数字来打扰你,只要你喜欢就添加(或删除)
数组:
public class LineUp {
Act[] acts;
int number;
public LineUp() {
acts = new Act[30]; // you assign to 'this' you don't create a new one
number = 0;
}
void addAct(Act a) {
acts[number] = a;
number++;
}
}
列表:
public class LineUp {
List<Act> acts;
public LineUp() {
acts = new ArrayList<>;
}
void addAct(Act a) {
acts.add(a);
}
}
所以这里我有一个 class 代表音乐节表演的表演。
public class Act {
private int num_members;
private String name;
private String kind;
private String stage;
public Act(int num_members, String name, String kind, String stage) {
this.num_members = num_members;
this.name = name;
this.kind = kind;
this.stage = stage;
}
public Act(int num_members, String name, String kind) {
this.num_members = num_members;
this.name = name;
this.kind = kind;
stage = null;
}
public int getNum_members() {
return num_members;
}
public void setNum_members(int num_members) {
this.num_members = num_members;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getKind() {
return kind;
}
public void setKind(String kind) {
this.kind = kind;
}
public String getStage() {
return stage;
}
public void setStage(String stage) {
this.stage = stage;
}
public String toString() {
return "(" + num_members + ", " + name + ", " + kind + ", " + stage + ")";
}
}
在另一个名为 LineUp 的 class 中,我想在一个名为 acts
的字段中存储最多 30 个行为,并且还有一个方法 addAct 将 Act
作为参数并将其添加到 acts
。这是我到目前为止使用的代码,但我不太确定该怎么做。
public class LineUp {
Act[] acts;
public LineUp() {
Act[] acts = new Act[30];
}
void addAct(Act a) {
}
}
如果您使用数组,您必须确保永远不需要添加超过 30 Act
并跟踪数字以帮助您,或者更简单的解决方案是使用 List<Act>
因为你不需要用数字来打扰你,只要你喜欢就添加(或删除)
数组:
public class LineUp {
Act[] acts;
int number;
public LineUp() {
acts = new Act[30]; // you assign to 'this' you don't create a new one
number = 0;
}
void addAct(Act a) {
acts[number] = a;
number++;
}
}
列表:
public class LineUp {
List<Act> acts;
public LineUp() {
acts = new ArrayList<>;
}
void addAct(Act a) {
acts.add(a);
}
}