在调用 this(arguments) 之前构造函数链接和准备参数
Constructor chaining and preparing arguments before calling this(aguments)
我正在制作 Yahtzee 游戏。我想为不同的情况提供一个构造函数。假设您懒得提供要创建新游戏的玩家的名字,我只想创建 "Unnamed Player 1"、"Unnamed Player 2" 等
以下是我尝试这样做的方法:
public class YahtzeeGame {
private List<Player> players = new ArrayList<>();
public YahtzeeGame(String[] playerNames) {
for (String playerName : playerNames) {
players.add(new Player(playerName));
}
}
public YahtzeeGame(int numberOfPlayers) {
String[] playerNames = new String[numberOfPlayers];
for (int i = 0; i < numberOfPlayers; i++) {
playerNames[i] = "Unnamed player " + (i+1);
}
this(playerNames); // ERROR: "Constructor call must be the first statement in a constructor.
}
public YahtzeeGame(String playerName) {
this(new String[] {playerName});
}
public YahtzeeGame() {
this("Unnamed player");
}
}
根据评论中的错误,这当然行不通。
有办法解决这个问题吗?为此我需要工厂模式吗?
是的,有一个相当简单的解决方法,至少在这种情况下是这样:创建一个将为您准备构造函数参数的静态方法。从 this
表达式调用它:
public YahtzeeGame(int numberOfPlayers) {
this(getUnnamedPlayers(numberOfPlayers));
}
private static String[] getUnnamedPlayers(int numberOfPlayers) {
String[] playerNames = new String[numberOfPlayers];
for (int i = 0; i < numberOfPlayers; i++) {
playerNames[i] = "Unnamed player " + (i+1);
}
return playerNames;
}
请注意,它必须是静态的,因为您也不能在链式构造函数之前调用 this
上的任何实例方法。
我正在制作 Yahtzee 游戏。我想为不同的情况提供一个构造函数。假设您懒得提供要创建新游戏的玩家的名字,我只想创建 "Unnamed Player 1"、"Unnamed Player 2" 等
以下是我尝试这样做的方法:
public class YahtzeeGame {
private List<Player> players = new ArrayList<>();
public YahtzeeGame(String[] playerNames) {
for (String playerName : playerNames) {
players.add(new Player(playerName));
}
}
public YahtzeeGame(int numberOfPlayers) {
String[] playerNames = new String[numberOfPlayers];
for (int i = 0; i < numberOfPlayers; i++) {
playerNames[i] = "Unnamed player " + (i+1);
}
this(playerNames); // ERROR: "Constructor call must be the first statement in a constructor.
}
public YahtzeeGame(String playerName) {
this(new String[] {playerName});
}
public YahtzeeGame() {
this("Unnamed player");
}
}
根据评论中的错误,这当然行不通。
有办法解决这个问题吗?为此我需要工厂模式吗?
是的,有一个相当简单的解决方法,至少在这种情况下是这样:创建一个将为您准备构造函数参数的静态方法。从 this
表达式调用它:
public YahtzeeGame(int numberOfPlayers) {
this(getUnnamedPlayers(numberOfPlayers));
}
private static String[] getUnnamedPlayers(int numberOfPlayers) {
String[] playerNames = new String[numberOfPlayers];
for (int i = 0; i < numberOfPlayers; i++) {
playerNames[i] = "Unnamed player " + (i+1);
}
return playerNames;
}
请注意,它必须是静态的,因为您也不能在链式构造函数之前调用 this
上的任何实例方法。