Question

我目前正在尝试建立一个简单的棋子运动程序。我从文本文件中读取了国际象棋运动表示法。国际象棋符号是这样的：

plb1 c3 =（p）pawn（l）白色从位置b1（b1）移动到位置c3（c3）。我的正则表达式然后解析移动。

在代码中我匹配各种部分并打印出各种动作。我想要做的是将我的 moveBetter（）方法分解为mutliple较小的方法，例如 findPiece（）或 findColor（）方法。

问题是，每当我尝试分解方法时，它们都不起作用，或者我被告知我必须使方法保持静态并进一步打破它。 如何将moveBetter（）拆分为多个方法？

import java.io.IOException;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class SuperiorFileIO {

public static void main(String[] args) throws IOException {
    moveBetter();
}

public static void moveBetter() throws IOException
{
    String piece = " "; //group1
    String color = " "; //group2
    String position = " "; //group3
    String position2 = " "; //group4
    String capture = " "; //group5
    //or 
    String kingPos1 = "";//6
    String kingPos2 = "";//7
    String rookPos1 = "";//8
    String rookPos2 = "";//9

    ReadFile readFile = new ReadFile();     
    String input = readFile.readTextFile("TestMoves");


    Pattern matchingMoves = Pattern.compile("(q|k|b|p|n|r+)(l|d)(\\w\\d) ?(\\w\\d)?([*])?|(\\w\\d) (\\w\\d) (\\w\\d) (\\w\\d)");
    Matcher m1 = matchingMoves.matcher(input);

    while (m1.find())
    {
        //find piece
            if(m1.group(6) != null && m1.group(6).matches("(\\w\\d)")) //castling part 1
                kingPos1 = m1.group(6);
            else if (m1.group(1).equals("q"))
                piece = "Queen";
            else if (m1.group(1).equals("k"))
                piece = "King";
            else if (m1.group(1).equals("b"))
                piece = "Bishop";
            else if (m1.group(1).equals("p"))
                piece = "Pawn";
            else if (m1.group(1).equals("n"))
                piece = "Knight";
            else if (m1.group(1).matches("(q|k|b|p|n|r+)") &&  m1.group(1).equals("r"))
                piece = "Rook";
            else {
                System.out.println("PIECE FAILED");
                piece = "piece Failed";
            }
        //find color
            if(m1.group(7) != null && m1.group(7).matches("(\\w\\d)")) //castling part 2
                kingPos2 = m1.group(7);
            else if(m1.group(2).equals("d"))
                color = "Black";
            else if(m1.group(2).matches("(l|d)") && m1.group(2).equals("l"))
                color = "White";
            else {
                System.out.println("COLOR FAILED");
                piece = "color Failed";
            }
        //set starting position
            if (m1.group(7) != null && m1.group(7).matches("(\\w\\d)") )
            rookPos1 = m1.group(7); //castling part 3
            else
                position = m1.group(3);
        //Castling
            if (m1.group(8) != null && m1.group(9) != null && m1.group(8).matches("(\\w\\d)") && m1.group(9).matches("(\\w\\d)"))
            {
                rookPos1 = m1.group(8);
                rookPos2 = m1.group(9);
                System.out.println("The king moves from " + kingPos1 + " to " + kingPos2 + " while the rook moves from " + rookPos1 + " to " + rookPos2);   
            }
        //if the player does not move // finds a piece
            else if (m1.group(4) == null)
                System.out.println("The " + color + " " + piece + " resides at " + position);           

        //set ending position
            else if (m1.group(4) != null && m1.group(5) == null)
            {
                position2 = m1.group(4);
                System.out.println("The " + color + " " + piece + " moves from " + position + " to " + position2);
            }
        //set ending position and capture
            else if (m1.group(4) != null && m1.group(5) != null)
            {
                //System.out.println("capture occured");
                position2 = m1.group(4);
                capture = m1.group(5); //unused
                System.out.println("The " + color + " " + piece + " moves from " + position + " to " + position2  + " and captures the piece on " + position2);
            }

            else {
                System.out.println("PARSE ERROR");
            }
    }


}

}

Answer 1

请勿使用静态movePiece方法致电main。定义一个ChessEngine类或其他东西，在main方法中实例化一个，然后让ChessEngine对象处理所有逻辑。然后，您可以按照其他答案/评论中的建议拆分您的方法。

Answer 2

你可以这样做：

public static void main(String[] args) throws IOException
{
    findPiece();
    findColor();
}

private static void findPiece()
{
    //Code
}

private static void findColor()
{
    //Code
}

方法可见性可以是私有的，也可以是公共的，具体取决于您希望如何使用它，我更倾向于私有这些方法。

findPiece（）：

while(m1.find())
{
    if(m1.group(6) != null && m1.group(6).matches("(\\w\\d)")) //castling part 1
        kingPos1 = m1.group(6);
    else if (m1.group(1).equals("q"))
        piece = "Queen";
    else if (m1.group(1).equals("k"))
        piece = "King";
    else if (m1.group(1).equals("b"))
        piece = "Bishop";
    else if (m1.group(1).equals("p"))
        piece = "Pawn";
    else if (m1.group(1).equals("n"))
        piece = "Knight";
    else if (m1.group(1).matches("(q|k|b|p|n|r+)") &&  m1.group(1).equals("r"))
        piece = "Rook";
    else {
        System.out.println("PIECE FAILED");
        piece = "piece Failed";
    }
}

对于其他方法几乎相同，循环和复制粘贴您要执行的部分。虽然如果在另一种方法中需要每种方法中的代码，那么拆分它将无法正常工作

Answer 3

考虑这个基本的例子：

public static void moveBetter() throws IOException
{
    String piece = "pluto";
    piece = "goofy";
}

为了将其拆分为两种方法，您必须：

1.提取变量并将其放在moveBetter外面

string piece;

public static void moveBetter() throws IOException
{
    piece = "pluto";
    piece = "goofy";
}

2.将变量定义为静态，因为静态方法（如moveBetter）无法访问非静态变量或非静态方法

static string piece;

3.创建一个新方法并将代码移入其中。记住新方法也必须是静态的，以便被moveBetter

访问

string static piece;

public static void moveBetter() throws IOException
{
    piece = "pluto";
    assignGoofy();
}

public static void assignGoofy()
{
    piece = "goofy";
}

将一种方法拆分为多种方法 - Java Chess

3 个答案: