Java 正在加载由int[]]中的数字定义的对象，isTopLeft（）不起作用

我正在通过一个2d数组加载游戏地图，其中每个数字都是10px。以下是一个文件示例：

"level_1_1.png", ?600:400?;
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0;
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0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0;
1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1;
1,1,2,2,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1;
1,1,2,2,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1;
1,1,2,2,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1;
1,1,2,2,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1;
1,1,2,2,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1;
0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0;
 0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0;

1是平台，2是玩家。我通过检查我是否在定义对象的左上角来加载它们（平台是20x20，因此是2x2正方形）。我通过倒数行和列来实现这一点，并按照预期的宽度（以十个像素为单位）修改该数字

以下是isTopLeft的代码：

    private boolean isTopLeft(int item, int r, int c, String[][] map){
    //width and height measured in # of map "steps" (10px ea)
    int width = 0;
    int height = 0;

    if(item == 0){

    }else if(item == 1){
        width = platformWidth/10;
        height = platformHeight/10;
    }else if(item == 2){
        width = playerWidth/10;
        height = playerHeight/10;
    }else if(item == 3){

    }else if(item == 4){

    }else if(item == 5){

    }else if(item == 6){

    }else if(item == 7){

    }else if(item == 8){

    }else if(item == 9){

    }

    //counts backwards and upwards to see how many blocks of the same type preceed it.
    int curR = r - 1;
    int prevWidthCount = 0;
    int prevHeightCount = 0;
    while(curR >= 0 && map[curR][c] == "" + item){
        curR -= 1;
        prevWidthCount += 1;
    }
    int curC = c - 1;
    while(curC >= 0 && map[r][curC] == "" + item){
        curC -= 1;
        prevHeightCount += 1;
    }


    if(prevWidthCount % width == 0 && prevHeightCount % height == 0){
        return true; 
    }

    return false;

}

---

有一件事突然向我袭来，那就是在台词中：

while(curR >= 0 && map[curR][c] == "" + item)

while(curC >= 0 && map[r][curC] == "" + item)

---

您正在使用

==

比较字符串。这是一个比较常见的错误，它确实会造成伤害-您必须使用

String.equals（）

来比较Java中的两个字符串。

不工作不是内置的错误消息。你得到了什么，你期望得到什么，为什么？我期望每2x2块1就放置一个平台。实际上，使用

==

非常适合插入字符串，也就是说它非常频繁地工作。这就是为什么它可能是一个如此隐蔽的错误。尽管我感谢您对最佳实践的关注。这不是问题所在。不过，谢谢：）这并不是最好的做法，更像是正确处理字符串，但如果您可以向isTopLeft提供一个示例输入，以及错误的输出（真/假），我可以在我的答案后面附加一些更具体的内容。@sdadas在使用

==

字符串比较以获得O（1）时，这听起来是一种非常好的调试方法绩效效益，故意的。不是有意使用它的新程序员通常不知道字符串的替代。这样，他们多次看到

==

有效，并假设它总是有效的，而实际上它只是有时有效。