【发布时间】:2018-11-04 06:32:49
【问题描述】:
我正在尝试通过我的 main() 函数传递此方法,但 loadMap 已经具有缓冲区读取器,因此我尝试使用它而不是创建自己的新缓冲区读取器。我该怎么做?
public static void main(String args[]) {
//throw exceptions here if args is empty
filename = args[0];
System.out.println(MapIO.loadMap(filename)[0]);
System.out.println(MapIO.loadMap(filename)[1]);
if (args.length < 1) {
System.err.println("Usage:\n" +"java CrawlGui mapname");
System.exit(1);
}
List<String> names=new LinkedList<String>();
try (BufferedReader reader = new BufferedReader(new FileReader(new
File(filename)))) {
String line;
while ((line = reader.readLine()) != null)
names.add(line);
System.out.println(names);
} catch (IOException e) {
e.printStackTrace();
}
MapIO.loadMap(filename);
launch(args);
}
/** Read information from a file created with saveMap
* @param filename Filename to read from
* @return null if unsucessful. If successful, an array of two Objects.
[0] being the Player object (if found) and
[1] being the start room.
* @detail. Do not add the player to the room they appear in, the caller
will be responsible for placing the player in the start room.
*/
public static Object[] loadMap(String filename) {
Player player = null;
try {
BufferedReader bf = new BufferedReader(
new FileReader(filename));
String line = bf.readLine();
int idcap = Integer.parseInt(line);
Room[] rooms = new Room[idcap];
for (int i = 0; i < idcap; ++i) {
line = bf.readLine();
if (line == null) {
return null;
}
rooms[i] = new Room(line);
}
for (int i = 0; i < idcap; ++i) { // for each room set up exits
line = bf.readLine();
int exitcount=Integer.parseInt(line);
for (int j=0; j < exitcount; ++j) {
line = bf.readLine();
if (line == null) {
return null;
}
int pos = line.indexOf(' ');
if (pos < 0) {
return null;
}
int target = Integer.parseInt(line.substring(0,pos));
String exname = line.substring(pos+1);
try {
rooms[i].addExit(exname, rooms[target]);
} catch (ExitExistsException e) {
return null;
} catch (NullRoomException e) {
return null;
}
}
}
for (int i = 0;i<idcap;++i) {
line = bf.readLine();
int itemcount = Integer.parseInt(line);
for (int j = 0; j < itemcount; ++j) {
line = bf.readLine();
if (line == null) {
return null;
}
Thing t = decodeThing(line, rooms[0]);
if (t == null) {
return null;
}
if (t instanceof Player) { // we don't add
player = (Player)t; // players to rooms
} else {
rooms[i].enter(t);
}
}
}
Object[] res = new Object[2];
res[0] = player;
res[1] = rooms[0];
return res;
} catch (IOException ex) {
return null;
} catch (IndexOutOfBoundsException ex) {
return null;
} catch (NumberFormatException nfe) {
return null;
}
}
【问题讨论】:
-
@James_D 当我的对象被返回时,我如何访问它们。这是 [0] 是 Player 对象(如果找到), [1] 是开始房间。现在,如果我打印它返回的结果 (@Room46362f) 和 (@Explorer343542) 相同。所以我知道它们是目标代码,但我想访问它们里面的内容
-
忍者,詹姆斯可能看不到此评论,因为您将其作为对您自己问题的评论而不是作为他回答下的评论。
标签: user-interface object javafx main