可绘制错误

Question

我为按钮创建了一个自定义可绘制对象，发现如果我在按钮数组之间共享该可绘制对象，那么我在按钮数组中按下的任何按钮都将导致该数组中的最后一个按钮也被按下。这是代码...

//xml code for stage button
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item android:state_pressed="true" 
android:drawable="@drawable/stage_button_selected"></item>
<item android:state_pressed="false" 
android:drawable="@drawable/stage_button_deselected"></item>
</selector>

//xml code for stage_button_selected
<layer-list xmlns:android="http://schemas.android.com/apk/res/android">
<item android:left="5dp" android:right="5dp" android:top="5dp" 
android:bottom="5dp">
<shape android:shape="oval" android:innerRadius="5dp">
<solid android:color="@android:color/holo_blue_light"></solid>
<stroke android:color="@android:color/holo_blue_dark" 
android:width="2dp"></stroke>
<corners android:radius="5dp"></corners>
</shape>
</item>
</layer-list>

//xml code for stage_button_deselected
<layer-list xmlns:android="http://schemas.android.com/apk/res/android">
<item android:left="5dp" android:right="5dp" android:top="5dp" 
android:bottom="5dp">
<shape android:shape="oval" android:innerRadius="5dp">
<solid android:color="@android:color/white"></solid>
<stroke android:color="#dddddd" android:width="2dp"></stroke>
<corners android:radius="5dp"></corners>
</shape>
</item>
</layer-list>`

//android code
Drawable drawable = resources.getDrawable(R.drawable.stage_button, null);
for(int ctr=0; ctr<btnStage.length; ctr++){
GridLayout.Spec specRow = GridLayout.spec(specStage[ctr][0], 1);
GridLayout.Spec specCol = GridLayout.spec(specStage[ctr][1], 1);
GridLayout.LayoutParams lp = new GridLayout.LayoutParams(specRow, 
specCol);
btnStage[ctr] = new Button(MainActivity.this);
btnStage[ctr].setBackground(drawable);
btnStage[ctr].setText("Stage " + stageNo(ctr));
btnStage[ctr].setWidth(btnWidth);
btnStage[ctr].setHeight(btnHeight);
final String temp = stageNo(ctr);
btnStage[ctr].setOnClickListener(MainActivity.this);
ss_gridLayout.addView(btnStage[ctr], lp);
}

该问题的解决方案是创建一个可绘制的数组以使用按钮数组，但是我不能在没有引起不想要的效果的情况下共享按钮数组之间的可绘制对象是没有意义的...