我可以使方法返回调用方方法吗？

我想干掉这段代码：

YourConsultant.GameState gameState() {    for (int i = 0; i < 3; i  ) {        // an 'xxx' column returns 'x', an 'ooo' returns 'o', a mixed row returns '0'        char c = areTheSame(board[i][0], board[i][1], board[i][2]);        if (c == 'x') {return YourConsultant.GameState.WON_BY_X;}        else if (c == 'o') {return YourConsultant.GameState.WON_BY_O;}    }    for (int i = 0; i < 3; i  ) {        char c = areTheSame(board[0][i], board[1][i], board[2][i]);        if (c == 'x') {return YourConsultant.GameState.WON_BY_X;}        else if (c == 'o') {return YourConsultant.GameState.WON_BY_O;}    }    {        char c = areTheSame(board[0][0], board[1][1], board[2][2]);        if (c == 'x') {return YourConsultant.GameState.WON_BY_X;}        else if (c == 'o') {return YourConsultant.GameState.WON_BY_O;}    }    {        char c = areTheSame(board[0][2], board[1][1], board[2][0]);        if (c == 'x') {return YourConsultant.GameState.WON_BY_X;}        else if (c == 'o') {return YourConsultant.GameState.WON_BY_O;}    }    ...}

为此,我想写一个简短的方法来做到这一点：

        if (c == 'x') {return YourConsultant.GameState.WON_BY_X;}        else if (c == 'o') {return YourConsultant.GameState.WON_BY_O;}

但这会使新方法返回.我想我不能做类似super.return的事情？我可以再次检查返回值,但不会使我的代码干燥.你有什么建议？ (对不起,如果以前有人问过,我发现这个很难搜索)

更新：我不能简单地传递值,因为如果areTheSame == 0,那么我不应该返回(还).

更新2：我修改了代码,用这两行替换每行：

        if (c == 'x' || c == 'o')  return declareWinner(c);

它工作正常,并且功能相同.仍然有一些重复,但更好的IMO.

解决方法:

不可以,方法不能为其调用者执行返回,但是调用者可以直接返回被调用方法返回的值.看来这不会满足您的目的,但是,因为您只想有条件地返回.

我会更深入地解决这个问题.请注意,您的四个节有多相似：有条件的收益不只是一点点潮湿.您想要执行的测试数量很少,无法枚举,因此您可以考虑以下方面的内容：

private final static int[][][] TRIPLES = new int[][][] {    { {0, 0}, {0, 1}, {0, 2} },    { {1, 0}, {1, 1}, {1, 2} },    { {2, 0}, {2, 1}, {2, 2} },    { {0, 0}, {1, 0}, {2, 0} },    { {0, 1}, {1, 1}, {2, 1} },    { {0, 2}, {1, 2}, {2, 2} },    { {0, 0}, {1, 1}, {2, 2} },    { {0, 2}, {1, 1}, {2, 0} },};YourConsultant.GameState gameState() {    for (int i = 0; i < TRIPLES.length; i  ) {        char c = areTheSame(            board[TRIPLES[i][0][0]][TRIPLES[i][0][1]],            board[TRIPLES[i][1][0]][TRIPLES[i][1][1]],            board[TRIPLES[i][2][0]][TRIPLES[i][2][1]]        );        if (c == 'x') {            return YourConsultant.GameState.WON_BY_X;        } else if (c == 'o') {            return YourConsultant.GameState.WON_BY_O;        }    }    return YourConsultant.GameState.NO_WINNER;}

来源：https://www.icode9.com/content-4-561851.html