我制作了一个有效的代码，但我的老师告诉我，我正在'发明'，因为他看到了很多if / else语句，我试着缩短我的代码，但我不知道如何摆脱这些如果/ else语句。
这是我的代码，简短摘要： 代码是一个nxn块的网格，只能与“空白”磁贴交换，所以我的代码看起来如下：
grd.setOnItemClickListener(new OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view,
int position, long id) {
// This statement prevents the player from playing the game
// if the timer is counting down.
if (timer == true) {
Toast.makeText(getApplicationContext(),
"The game has not started yet!", Toast.LENGTH_SHORT)
.show();
} else {
clickcounter++;
// Because difficulty's are 0,1,2 the row/column dimensions
// are difficulty + 3, here i create the texview of the moves
// as well.
size = (int) difficulty + 3;
TextView movesText = (TextView) findViewById(R.id.textView2);
// Checks if click is even or odd. Even is firstclick odd is
// secondclick
if (clickcounter % 2 == 1) {
firstimg = (ImageView) view;
firstclick = position;
firsttag = (int) firstimg.getTag();
} else if (clickcounter % 2 == 0) {
ImageView secondimg = (ImageView) view;
scndtag = (int) secondimg.getTag();
// To prevent cheaters to press twice on the same tile
// we set this if statement
if (position == firstclick) {
clickcounter = 1;
// Next we check whether one of the clicked images
// is the blank one, if that's not the case do
// nothing!
} else if (firsttag != crops.size() - 1
&& scndtag != crops.size() - 1) {
// This else statements checks what happens if one has clicked the
// blank tile
} else {
// This statement below is the same as an switch/case statement
// but because of the fact that the switch/case statement gave the
// following error i chose for this method:
// case expressions must be constant expressions
int i = firstclick - position;
if (i == -1) {
if(firstclick % size == size - 1){
} else {
swap(firstclick,position);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
}
} else if (i == 1) {
if(firstclick % size == 0){
} else {
swap(firstclick,position);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
}
} else if (i == size) {
swap(firstclick,position);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
} else if (i == -size) {
swap(firstclick,position);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
}
}
}
}
}
});
}
我试图制作一个switch / case语句以防止if / else语句的总和，但是它说它无法将大小标识为常量（因为难度是玩家选择的变量），所以如果我能设法做到这一点，if / else语句的重载可能会消失。
有什么建议吗？我已经做了一些工作来使代码更具可读性，原始代码如下：
grd.setOnItemClickListener(new OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view,
int position, long id) {
if (timer == true) {
Toast toast = new Toast(getApplicationContext());
toast.makeText(getApplicationContext(),
"The game has not started yet!", Toast.LENGTH_SHORT)
.show();
} else {
clickcounter++;
// Because difficulty's are 0,1,2 the row/column dimensions
// are difficulty + 3
size = (int) difficulty + 3;
TextView movesText = (TextView) findViewById(R.id.textView2);
// Checks if click is even or odd. Even is firstclick odd is
// secondclick
if (clickcounter % 2 == 1) {
firstimg = (ImageView) view;
firstclick = position;
firsttag = (int) firstimg.getTag();
} else if (clickcounter % 2 == 0) {
ImageView secondimg = (ImageView) view;
Bitmap swapImage = crops.get(position);
Integer swapID = ID.get(position);
scndtag = (int) secondimg.getTag();
// To prevent cheaters to press twice on the same tile
// we set this if statement
if (position == firstclick) {
clickcounter = 1;
// These next if/else if statements are a summation
// of all kinds of
// rules that need to be implemented in the game!
// First we check whether one of the clicked images
// is the blank one
} else if (firsttag != crops.size() - 1
&& scndtag != crops.size() - 1) {
// Here we check if the tiles are neighbors of each
// other
} else if ((position - 1 != firstclick)
&& (position + 1 != firstclick)
&& (position + size) != firstclick
&& ((position - size) != firstclick)) {
// Check when secondclick on the right of firstclick
} else if (position - 1 == firstclick) {
// This prevents the game to swap tiles on the
// edges, like
// starting from 0, 2 with 3.
if (position % size != 0) {
crops.set(position, crops.get(firstclick));
ID.set(position, ID.get(firstclick));
crops.set(firstclick, swapImage);
ID.set(firstclick, swapID);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
}
// Check when secondclick on the left of firstclick
} else if (position + 1 == firstclick) {
if (position % size != (size - 1)) {
crops.set(position, crops.get(firstclick));
ID.set(position, ID.get(firstclick));
crops.set(firstclick, swapImage);
ID.set(firstclick, swapID);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
}
// Check when secondclick is above first
} else if ((position + size) == firstclick) {
crops.set(position, crops.get(firstclick));
ID.set(position, ID.get(firstclick));
crops.set(firstclick, swapImage);
ID.set(firstclick, swapID);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
// Check when secondclick is above first
} else if ((position - size) == firstclick) {
crops.set(position, crops.get(firstclick));
ID.set(position, ID.get(firstclick));
crops.set(firstclick, swapImage);
ID.set(firstclick, swapID);
moves++;
movesText.setText("Moves: " + moves);
checkwin2();
}
// This useful command rebuilds the imageviews
grd.invalidateViews();
}
}
}
});
}
亲切的问候，
Kees Til
答案 0 :(得分：2)
首先，＆＃34;嵌套过多的固有问题＆＃34;是代码维护。因此，学习如何减少它是一种好习惯。
当你有嵌套循环/ if-else时，你可以做一些事情来试图消除它。以下是一些指示：
如果你在代码中看到重复的模式，在你的情况下是这样的：
交换（firstclick，位置）; 移动++; movesText.setText（＆＃34; Moves：＆＃34; + move）; checkwin2（）;
您应该创建一个方法。然后，如果由于任何原因它改变了，它就在一个地方。此外，这允许您分离＆＃34;逻辑＆＃34; （if-else）来自程序。
您的代码非常＆＃34;程序性＆＃34;这有时是必要的，但却失去了面向对象编程的好处。要更改此设置，请考虑创建一个对象来处理此问题。制作＆＃34;逻辑变量＆＃34;像这样：
int i = firstclick - position;
并使用此示例，make＆＃34; firstclick＆＃34;和＆＃34;位置＆＃34;进入私人对象属性并创建＆＃34; getters＆＃34;和＃34; setters＆＃34;。然后根据get / set执行逻辑。这将使您能够专注于对象＆＃34;受到影响而受到影响。然后，如果您执行了第1步，则将这些方法合并到新对象中。
此外，您可以尝试采用最深层的嵌套逻辑，将其转换为＆＃34;对象＆＃34;然后尝试对下一个最深处做同样的事情，依此类推。这可以揭示不明显的逻辑模式，同时也有助于使代码更易于阅读和维护。
检查你的＆＃34;标志＆＃34;变量。也许他们应该被分成一个或多个对象。您甚至可以将对象用作＆＃34;标记＆＃34; - 甚至标志可以成为代码中的两个或多个变量，而不是正确地成为单个对象。
注意：不要害怕＆＃34;害怕＆＃34;只有一种或两种方法的小型或简单物体。此外，不要犹豫，有单行方法。重要的是动作（方法）和属性（变量）放在对象中，因为它们是相关的。